H  Y  D  R  A  U 


BY 

HORACE  W.   KING 

Professor  of  Hydraulic   Engineering, 
University  of  Michigan 

AND 

CHESTER  O.  WISLER 

Associate  Professor  of  Hydraulic  Engineering, 
University  of  Michigan 


NEW  YORK 

JOHN  WILEY   &  SONS,  INC 

LONDON:  CHAPMAN  &  HALL,  LIMITED 
1922 


••-:..:/••.••• 


Copyright,  1922, 
By  HORACE  W.  KING 

AND 

CHESTER  O.  WISLER 


PRESS  OF 

BRAUNWORTH  &  CO 

BOOK  MANUFACTURERS 

BROOKLYN,    N.    Y. 


PREFACE 


THIS  book  deals  with  the  fundamental  principles  of  hydraulics 
and  their  application  in  engineering  practice.  Though  many 
formulas  applicable  to  different  types  of  problems  are  given,  it 
has  been  the  aim  of  the  authors  to  bring  out  clearly  and  logically, 
the  underlying  principles  which  form  the  basis  of  such  formulas 
rather  than  to  emphasize  the  importance  of  the  formulas  them- 
selves. 

Our  present  knowledge  of  fluid  friction  has  been  derived 
largely  through  experimental  investigation  and  this  has  resulted 
in  the  development  of  a  large  number  of  empirical  formulas. 
Many  of  these  formulas  have  necessarily  been  included  but, 
in  so  far  as  possible,  the  base  formulas  to  which  empirical  coeffi- 
cients have  been  applied  have  been  derived  analytically  from 
fundamental  consideration  of  basic  principles. 

The  book  is  designed  as  a  text  for  beginning  courses  in 
hydraulics  and  as  a  reference  book  for  engineers  who  may  be 
interested  in  the  fundamental  principles  of  the  subject.  Tables 
of  coefficients  are  given  which  are  sufficiently  complete  for  class- 
room work,  but  the  engineer  in  practice  will  need  to  supplement 
them  with  the  results  of  his  own  experience  and  with  data  obtained 
from  other  published  sources. 

Chapters  I  to  VI  inclusive  and  Chapter  XI  were  written  by 
Professor  Wisler  and  Chapters  VII  to  X  inclusive  were  written 
by  Professor  King.  Acknowldgement  for  material  taken  from 
many  publications  is  made  at  the  proper  place  in  the  text. 

H.  W.  K. 
C.  O.  W. 

University  of  Michigan, 
April,  1922. 

5019i 3 

ill 


CONTENTS 


CHAPTER  I 

INTRODUCTION 

PAGE 

ART.  1. — Fluids.  2.  Definitions.  3.  Units  used  in  Hydraulics.  4. 
Weight  of  Water.  5.  Compressibility  of  Water.  6.  Viscosity.  7. 
Surface  Tension.  8.  Accuracy  of  Computations 1 


CHAPTER   II 
PRINCIPLES  OF  HYDROSTATIC  PRESSURE 

ART.  9. — Intensity  of  Pressure.  10.  Direction  of  Resultant  Pressure. 
11.  Pascal's  Law.  12.  Free  Surface  of  a  Liquid.  13.  Atmospheric 
Pressure.  14.  Vacuum.  15.  Absolute  and  Gage  Pressure.  16. 
Intensity  of  Pressure  at  any  Point.  17.  Pressure  Head.  18.  Trans- 
mission of  Pressure.  19.  Vapor  Pressure.  20.  The  Mercury 
Barometer.  21.  Piezometer  Tubes.  22.  Mercury  Gage.  23.  The 
Differential  Gage,  24.  Suction  Pumps  and  Siphons , 7 

CHAPTER  III 

PRESSURE  ON  SURFACES 

ART.  25. — Total  Pressure  on  Plane  Areas.  26.  Center  of  Pressure  on 
Plane  Areas.  27.  Graphical  Method  of  Location  of  Center  of 
Pressure.  28.  Position  of  Center  of  Pressure  with  Respect  to 
Center  of  Gravity.  29.  Horizontal  and  Vertical  Components  of 
Pressure  on  any  Surface 2o 

CHAPTER  IV 
IMMERSED  AND  FLOATING  BODIES 

ART.  30. — Principle  of  Archimedes.  31.  Center  of  Buoyancy.  32.  Sta- 
bility of  Floating  Bodies.  33.  Determination  of  Metacentric 
Height 38 


Vl  CONTENTS 

CHAPTER  V 

RELATIVE  EQUILIBRIUM  OF  LIQUIDS 

PAGE 

ART.  34. — Relative  Equilibrium  Defined.  35.  Vessel  Moving  with 
Constant  Linear  Acceleration.  36.  Vessel  Rotating  about  a  Vertical 
Axis 45 

CHAPTER  VI 

PRINCIPLES  OF  HYDROKINETICS 

ART.  37. — Introductory.  38.  Friction.  39.  Discharge.  40.  Steady 
Flow  and  Uniform  Flow.  41.' Continuity  of  Discharge.  42.  Stream 
Line  and  Turbulent  Motion.  43.  Energy  and  Head.  44.  Ber- 
noulli's Theorem.  45.  Application  of  Bernoulli's  Theorem  to 
Hydrostatics.  46.  Bernoulli's  Theorem  in  Practice.  47.  Venturi 
Meter,  48,  Pitot  Tube 51 

CHAPTER  VII 
FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 

ART.  49. — Description  and  Definitions.  50.  Characteristics  of  the  Jet. 
51.  Fundamental  Orifice  Formula.  52.  Orifice  Coefficients.  53. 
Algebraic  Transformation  of  Formula  with  Velocity  of  Approach 
Correction.  54.  Head  Lost  in  an  Orifice.  55.  Path  of  Jet.  56. 
Orifices  under  Low  Heads.  57.  Suppression  of  Contraction.  58. 
Standard  Short  Tube.  59.  Converging  Tubes.  60.  Nozzles.  61. 
Diverging  Tubes.  62.  Borda's  Mouthpiece.  63.  Re-entrant  Tubes. 
64.  Submerged  Orifice.  65.  Partially  Submerged  Orifices.  66. 
Gates.  67.  Discharge  under  Falling  Head 71 

CHAPTER  VIII 
FLOW  OF  WATER  OVER  WEIRS 

ART.  68. — Description  and  Definitions.  69.  Velocity  at  any  Depth. 
70.  Theoretical  Formulas  for  Discharge.  71.  Theoretical  Formula 
for  Mean  Velocity.  72.  Weir  Coefficients.  73.  Weirs  with  End 
Contractions.  74.  Modifications  of  Fundamental  Formula.  75. 
Algebraic  Transformation  of  Formula.  76.  Weir  Experiments. 
77.  Formulas  for  Sharp-crested  Weirs.  78.  Discussion  of  Weir 
Formulas.  79.  Submerged  Weirs.  80.  Further  Discussion  of  Sub- 
merged Weirs.  81.  Triangular  Weirs.  82.  Trapezoidal  Weirs. 
83.  The  Cippoletti  Weir.  84.  Weirs  not  Sharp-crssted.  85.  Broad- 
crested  Weirs  or  Chutes.  86.  Measurement  of  Head.  87.  Con- 
ditions for  Accurate  Measurement  over  Sharp-crested  Weirs 101 


CONTENTS  Vii 

CHAPTER  IX 

FLOW  OF  WATER  THROUGH  PIPES 

PAGE 

ART.  88. — Description  and  Definitions.  89.  Wetted  Perimeter  and 
Hydraulic  Radius.  90.  Critical  Velocities  in  Pipes.  91.  Friction 
and  Distribution  of  Velocities.  92.  Energy  of  Water  in  a  Pipe.  93. 
Continuity  of  Flow  in  Pipes.  94.  Loss  of  Head.  95.  Hydraulic 
Gradient.  96.  Loss  of  Head  Due  to  Friction  in  Pipes.  97.  The 
Chezy  Formula.  98.  Hazen- Williams  Formula.  99.  King  Formula. 
100.  General  Discussion  of  Pipe  Formulas.  101.  Friction  Formula 
for  Non-turbulent  Flow.  102.  Detailed  Study  of  Hydraulic  Gra- 
dient and  Minor  Losses.  103.  Part  of  Pipe  above  Hydraulic  Gra- 
dient. 104.  Special  Problems.  105.  Branching  Pipe  Connecting 
Reservoirs  at  Different  Elevations.  '106.  Compound  Pipe  Con- 
necting Two  Reservoirs.  107.  Pipes  of  More  than  One  Diameter 
Connected  in  Series 135 

CHAPTER  X 
FLOW  OF  WATER  IN  OPEN  CHANNELS 

ART.  108. — Description  and  Definition.  109.  Wetted  Perimeter  and 
Hydraulic  Radius.  110.  Friction  and  Distribution  of  Velocities. 
111.  Energy  Contained  in  Water  in  an  Open  Channel.  112.  Con- 
tinuity of  Flow  in  Open  Channels.  113.  Loss  of  Head.  114.  Hy- 
draulic Gradient  or  Water  Surface.  115.  Loss  of  Head  Due  to 
Friction  in  Open  Channels.  116.  The  Chezy  Formula.  117.  The 
Kutter  Formula.  118.  The  Manning  Formula.  119.  Comparison 
of  Manning  and  Kutter  Formulas.  120.  The  Bazin  Formula.  121. 
Open-channel  Formulas  in  General.  122.  Detailed  Study  of  Hy- 
draulic Gradient  or  Water  Surface.  123.  Hydraulics  of  Rivers. 
124.  Irregular  Sections.  125.  Cross-section  of  Greatest  Efficiency. 
126.  Circular  Sections.  127.  Non-uniform  Flow.  128.  Backwater. 
129.  Divided  Flow 176 

CHAPTER  XI 
HYDRODYNAMICS 

ART.  130. — Fundamental  Principles.  131.  Interpretation  of  Newton's 
Laws.  132.  Relative  and  Absolute  Velocities.  133.  Jet  Impinging 
Normally  on  a  Fixed  Flat  Plate.  134.  Jet  Impinging  Normally 
on  a  Moving  Flat  Plate.  135.  Jet  Deflected  by  a  Fixed  Curved 
Vane.  136.  Jet  Deflected  by  a  Moving  Curved  Vane.  137.  Work 
Done  on  Moving  Vanes.  138.  Forces  Exerted  upon  Pipes.  139. 
Straight  Pipe  of  Varying  Diameter.  140.  Pipe  Bends.  141.  Water 
Hammer  in  Pipe  Lines.  142.  Formulas  for  Water  Hammer 211 


HYDRAULICS 


CHAPTER   I 
INTRODUCTION 

1.  Fluids. — Fluids  are  substances  which  possess  unlimited 
mobility  and  which  offer  practically  no  resistance  to  change  of 
form.     A  perfect  fluid  yields  to  the  slightest  tangential  stress, 
and  can  therefore  have  no  tangential  stress  if  it  is  at  rest.   Fluids 
may  be  divided  into  two  classes,  (a)  liquids,  or  fluids  that  are 
practically  incompressible,  and  (6)  gases,  or  fluids  that  are  highly 
compressible. 

2.  Definitions. —  Hydraulics   is   the   science   embodying  the 
laws  that  relate  to  the  behavior  of  liquids,  and  particularly  of 
water.     In  its  original  sense  the  term  hydraulics  was  applied 
only  to  the  flow  of  water  in  conduits,  but  the  scope  of  the  word 
has  been  broadened  by  usage. 

Hydraulics  may  be  divided  conveniently  into  three  branches : 
(a)  hydrostatics,  which  deals  with  liquids  at  rest,  (6)  hydrokinetics, 
which  treats  of  the  laws  governing  the  flow  of  liquids,  and  (c) 
hydrodynamics,  which  relates  to  the  forces  exerted  upon  other 
objects  by  liquids  in  motion  or  upon  liquids  by  other  objects  in 
motion. 

The  fundamental  laws  of  hydraulics  apply  equally  to  all 
liquids,  but  in  hydrokinetics  empirical  coefficients  must  be  mod- 
ified to  conform  to  the  liquid  considered.  Water  is  the  most 
common  liquid  and  the  only  one  that  is  of  general  interest  to 
engineers. 

3.  Units  Used  in  Hydraulics. — It  is  common  practice  in  the 
United  States  and  Great  Britain  to  base  hydraulic  computations 
on  the  foot-pound-second  system  of  units.     In  practically  all 
hydraulic  formulas  these  units  are  used,  and  if  not  otherwise 


INTRODUCTION 


stated  they  are  understood.  Frequently  the  diameters  of  pipes 
or  orifices  are  expressed  in  inches,  pressures  are  usually  stated  in 
pounds  per  square  inch,  and  volumes  may  be  expressed  in  gallons. 
Before  applying  such  data  to  problems,  conversion  to  the  foot- 
pound-second system  of  units  should  be  made.  Care  in  the 
conversion  of  units  is  essential.  Errors  in  hydraulic  computa- 
tions result  more  frequently  from  wrong  use  of  units  than  from 
any  other  cause. 

4.  Weight  of  Water. — Water  has  its  maximum  density  at  a 
temperature  of  39.3°  F.  At  this  temperature  pure  water  has 
been  given  a  specific  gravity  of  unity  and  it  thus  serves  as  a 
standard  of  density  for  all  substances.  The  density  of  water 
decreases  for  temperatures  above  and  below  39.3°.  It  freezes 
at  32°  and  boils  at  212°  F.  The  weight  of  pure  water  at  its 
temperature  of  maximum  density  is  62.424  Ibs.  per  cubic  foot. 
The  weights  at  various  temperatures  are  given  in  the  following 
table: 

WEIGHT  OF  PURE  WATER 


Tempera- 

Pounds 

Tempera- 

Pounds 

Tempera- 

Pounds 

ture, 

per 

ture, 

per 

ture, 

per 

Fahrenheit 

cubic  foot 

Fahrenheit 

cubic  foot 

Fahrenheit 

cubic  foot 

32° 

62.416 

90° 

62.118 

160° 

61.006 

39.3 

62.424 

100 

61.998 

170 

60.799 

40 

62.423 

110 

61.865 

180 

60.586 

50 

62  .  408 

120 

61.719 

190 

60.365 

60 

62.366 

130 

61.555 

200 

60.135 

70 

62.300 

140 

61.386 

210 

59.893 

80 

62.217 

150 

61  .  203 

212 

59.843 

For  temperatures  above  the  boiling  point  Rankine  gives  the 
following  approximate  formula:  w  being  the  weight  of  water  in 
pounds  per  cubic  foot  and  T  the  temperature  in  degrees  Fahren- 
heit, 

124.85 


w  = 


T+461  ,      500 


500     '  T+4G1 


As  water  occurs  in  nature,  it  invariably  contains  a  certain 
amount  of  salts  and  mineral  matter  in  solution.     Silt  or  other 


COMPRESSIBILITY  OF  WATER 


3 


impurities  may  also  be  carried  in  suspension.  These  substances 
are  invariably  heavier  than  water  and  they  therefore  increase  its 
weight.  The  impurities  contained  in  rivers,  inland  lakes  and 
ordinary  ground  waters  do  not  usually  add  more  than  one-tenth 
of  a  pound  to  the  weight  per  cubic  foot.  Ocean  water  weighs 
about  64  Ibs.  per  cubic  foot.  After  long-continued  droughts  the 
waters  of  Great  Salt  Lake  and  of  the  Dead  Sea  have  been  found 
to  weigh  as  much  as  75  Ibs.  per  cubic  foot. 

Since  the  weight  of  inland  water  is  not  greatly  affected  by 
ordinary  impurities  nor  changes  of  temperature,  an  average 
weight  of  water  may  be  used  which  usually  will  be  close  enough 
for  hydraulic  computations.  In  this  book  the  weight  of  a  cubic 
foot  of  water  is  taken  as  62.4  Ibs.  Sea  water  will  be  assumed 
to  weigh  64.0  Ibs.  per  cubic  foot  unless  otherwise  specified.  In 
very  precise  work  weights  corresponding  to  different  temperatures 
may  be  taken  from  the  above  table. 

6.  Compressibility  of  Water. — Water  is  commonly  assumed  to 
be  incompressible,  but  in  reality  it  is  slightly  compressible.  Upon 
release  from  pressure  water  immediately  regains  its  original  volume. 
For  ordinary  pressures  the  modulus  of  elasticity  is  constant— 
that  is,  the  amount  of  compression  is  directly  proportional  to  the 
pressure  applied.  The  modulus  of  elasticity,  E,  varies  with  the 
temperature  as  shown  in  the  following  table. 


Temperature, 
Fahrenheit 

Modulus  of  elasticity, 
pounds  per  square  inch 

35° 
77 
212 

288,000 
327,000 
360,000 

These  values  hold  only  for  pressures  below  1000  Ibs.  per 
square  inch.  Kite  obtained  a  reduction  in  volume  of  10  per  cent 
for  a  pressure  of  65,000  Ibs.  per  square  inch,  giving  a  value  of  E 
of  650,000  for  this  high  intensity  of  pressure. 

The  compressibility  of  water  usually  affects  the  solution  of 
practical  problems  in  hydraulics  only  by  changing  its  unit  weight. 
Since  pressures  commonly  encountered  are  relatively  small,  in 
most  cases  water  may  be  considered  incompressible  without 
introducing  any  appreciable  error. 


4  INTRODUCTION 

6.  Viscosity. — One  of  the  characteristic  properties  of  a  liquid 
is  its  ability  to  flow.     A  perfect  liquid  would  be  one  in  which  every 
particle  could  move  in  contact  with  adjacent  particles  without 
friction.     The  pressures  between  all  such  particles  would  be  normal 
to  their  respective  surfaces  at  the  points  of  contact  since  there 
could  be  no  tangential  stress  without  friction.     All  liquids  are 
capable,  however,  of   resisting   a   certain   amount   of  tangential 
stress,  and  the  extent  to  which  they  possess  this  ability  is  a 
measure  of  their  viscosity. 

Water  is  one  of  the  least  viscous  of  all  liquids.     Oil,  molasses 
and  wax  are  examples  of  liquids  having  greater  viscosity. 

7.  Surface  Tension. — At  any  point  within  a  body  of  liquid 
the  molecules  are  attracted  towards  each  other  by  equal  forces. 
The  molecules  forming  the  surface  layer,  however,  are  subjected 
to  an  attraction  downward  that  is  not  balanced  by  an  upward 
attraction.     This  causes  a  film  or  skin  to  form  on  the  surface  and 
results  in  many  interesting  phenomena.     A  needle  may  be  made  to 
float  upon  water  so  long  as  the  surface  film  is  not  broken,  but  it 
will  sink  immediately  when  the  film  is  broken.     Surface  tension 
causes  the  spherical  shape  of  dewdrops  or  drops  of  rain.     This 
phenomenon   also  makes   possible   the  hook  gage   described   in 
Art.  86.     Where  water  flows  in  an  open  conduit,  surface  tension 
retards  velocities  at  the  surface,  the  maximum  velocity  ordinarily 
being  below  the  surface  (see  Art.  110).     Capillary  action  is  also 
explained  by  the  phenomenon  of  surface  tension  combined  with 
that  of  adhesion. 

A  (Fig,  1)  illustrates  an  open  tube  of  small  diameter  im- 
mersed in  a  liquid  that  wets 
the  tube.  Water  rises  in  the 
tube  higher  than  the  level  out  - 
side,  the  meniscus  being  con- 
cave upward.  The  tube  B  is 
immersed  in  mercury  or  some 

FlG  j  other  liquid    which   does  not 

wet  the  tube.     In  this  case  the 

meniscus  is  convex  upward  and  the  level  of  the  liquid  in  the  tube 
is  depressed.  The  effect  of  capillarity  decreases  as  the  size  of 
tube  increases.  The  water  in  a  tube  one-half  inch  in  diameter 
is  approximately  at  the  same  level  as  the  outside  water,  but  it 
is  appreciably  different  for  smaller  tubes.  For  this  reason, 


ACCURACY   OF   COMPUTATIONS  5 

piezometer  tubes  (Art.  21)  should  not  have  diameters  much 
smaller  than  one-half  inch. 

8.  Accuracy  of  Computations.  —  Accuracy  of  computations  is 
most  desirable,  but  results  should  not  be  carried  out  to  a  greater 
number  of  significant  figures  than  the  data  justify.  Doing  this 
implies  an  accuracy  which  does  not  exist  and  may  give  results 
that  are  entirely  misleading. 

Suppose,  for  example,  that  it  be  desired  to  determine  the 
theoretical  horse-power  available  in  a  stream  where  the  discharge 
is  311  cu.  ft.  per  second  and  the  available  head  is  12.0  ft.  The 
formula  to  be  used  is 

_wQH 


where    w  =  the  weight  of  a  cubic  foot  of  water; 

Q  =  the  discharge  in  cubic  feet  per  second; 
77  =  the  available  head. 


Substituting  in  the  formula, 
62.4X311X12.0 


550 


=  423.41  horse-power. 


Referring  to  t}ie  table,  page  2,  it  is  seen  that  for  the  ordinary 
range  of  temperatures  the  weight  of  water  may  vary  from  62.30 
to  62.42  Ibs.  per  cubic  foot.  Furthermore,  the  statement  that  the 
discharge  is  311  cu.  ft.  per  second  means  merely  that  the  exact 
value  is  more  nearly  311  than  310  or  312.  In  other  words,  the 
true  value  may  lie  anywhere  between  310.5  and  311.5.  Likewise, 
the  fact  that  the  head  is  given  as  12.0  merely  indicates  that  the 
correct  value  lies  somewhere  between  11.95  and  12.05.  Therefore 
substituting  in  the  formula  the  lower  of  these  values, 

62.30X310n5Xll.95  =  42030  horse.power. 
550 

Again  substituting  in  the  formula  the  higher  of  the  possible 
values, 

62.42X311.5X12.05 


550 


=  426 . 00  horse-power. 


It  is  evident,  therefore,  that  the  decimal  .41  in  the  original 
answer  423.41  is  unjustified,  and  that  the  last  whole  number,  3, 


6  INTRODUCTION 

merely  represents  the  most  probable  value,  since  the  correct 
value  may  lie  anywhere  between  420.30  and  426.00.  The  answer 
should,  therefore,  be  given  as  423. 

It  may  be  stated  in  general  that  in  any  computation  involving 
multiplication  or  division,  in  which  one  or  more  of  the  numbers  is 
the  result  of  observation,  the  answer  should  contain  the  same 
number  of  significant  figures  as  is  contained  in  the  observed  quan- 
tity having  the  fewest  significant  figures.  In  applying  this  rule  it 
should  be  understood  that  the  last  significant  figure  in  the  answer 
is  not  necessarily  correct,  but  represents  merely  the  most  probable 
value.  To  give  in  the  answer  a  greater  number  of  significant 
figures  indicates  a  degree  of  accuracy  that  is  unwarranted  and 
misleading. 


CHAPTER  II 
PRINCIPLES   OF  HYDROSTATIC  PRESSURE 

9.  Intensity  of  Pressure. — The  intensity  of  pressure  at  any 
point  in  a  liquid  is  the  amount  of  pressure  per  unit  area. 

If  the  intensity  of  pressure  is  the  same  at  every  point  on  any 
area,  A, 

,-£  =  £ CD 

the  symbol  p  representing  the  intensity  of  pressure  and  P  the 
total  pressure  acting  upon  the  area. 

If,  however,  the  intensity  of  pressure  is  different  at  different 
points,  the  intensity  of  pressure  at  any  point  will  be  equal  to  the 
pressure  on  a  small  differential  area  surrounding  the  point  divided 
by  the  differential  area,  or 

P-£,  «> 


Intensities  of  pressure  are  commonly  expressed  in  pounds 

per  square  inch  and  pounds  per    =  ^^-^^^^ __^ 

square  foot.  Where  there  is  no 
danger  of  ambiguity,  the  term 
pressure  is  often  used  as  an  abbre- 
viated expression  for  intensity  of 
pressure. 

10.  Direction  of  Resultant 
Pressure. — The  resultant  pressure 
on  any  plane  in  a  liquid  at  rest  is 
normal  to  that  plane. 

Assume  that  the  resultant  pres- 
sure P,  on  any  plane  AB  (Fig.  2), 
makes  an  angle  other  than  90°  with  the  plane.  Resolving  P 


FIG.  2. 


into    rectangular    components  PI 

7 


and   P2,  respectively   parallel 


8  PRINCIPLES  OF  HYDROSTATIC  PRESSURE 

with  and  perpendicular  to  AB,  gives  a  component  PI  which  can 
be  resisted  only  by  a  tangential  stress.  By  definition,  a  liquid  at 
rest  can  resist  no  tangential  stress  and  therefore  the  pressure 
must  be  normal  to  the  plane.  This  means  that  there  can  be  no 
static  friction  in  hydraulics. 

11.  Pascal's  Law. — At  any  point  within  a  liquid  at  rest,  the 
intensity  of  pressure  is  the  same  in  all  directions.  This  principle 
is  known  as  Pascal's  Law. 

Consider  an  infinitesimally  small  wedge-shaped  volume,  BCD 
(Fig,  3),  in  which  the  side  BC  is  vertical,  CD  is  horizontal  and 

BD  makes  any  angle  6  with  the  hori- 
zontal. Let  AI,  AZ  and  AS  and  pi, 
P2  and  ps  represent,  respectively  the 
areas  of  these  sides  and  the  intensities 
of  pressure  to  which  they  are  subjected. 
Assume  that  the  ends  of  the  wedge  are 
"*  vertical  and  parallel. 

Since  the  wedge  is  at  the  rest,  the 
principles  of  equilibrium  may  be  ap- 
plied to  it.     From  Art.  10  it  is  known 
F  that  the  pressures  are  normal  to  the 

faces   of   the   wedge.      Choosing    the 

coordinate  axes  as  indicated  in  Fig.  3  and  setting  up  the  equations 
of  equilibrium,  SY  =  0  and  SF  =  0,  and  neglecting  the  pressures 
on  the  ends  of  the  wedge,  since  they  are  the  only  forces  acting 
on  the  wedge  which  have  components  along  the  Z-axis  and 
therefore  balance  each  other,  the  following  expressions  result; 


in  6, 

m.    .  cos  0- 
But 

As  sin  6= AI     and    As  cos  6 
Therefore 


Since  BD  represents  a  plane  making  any  angle  with  the  hori- 
zontal and  the  wedge  is  infinitesimally  small  so  that  the  sides  may 
be  considered  as  bounding  a  point,  it  is  evident  that  the  intensity 
of  pressure  at  any  point  must  be  the  same  in  all  directions. 


FREE  SURFACE  OF  A  LIQUID 


9 


12.  Free  Surface  of  a  Liquid. — Strictly  speaking,  a  liquid 
having  a  free  surface  is  one  on  whose  surface  there  is  absolutely 
no  pressure.     It  will  be  shown  later,  however,  that  there  is  always 
some  pressure  on  the  surface  of  every  liquid. 

In  practice  the  free  surface  of  a  liquid  is  considered  to  be  a 
surface  that  is  not  in  contact  with  the  cover  of  the  containing 
vessel.  Such  a  surface  may  or  may  not  be  subjected  to  the  pres- 
sure of  the  atmosphere. 

It  may  be  shown  that  the  free  surface  of  a  liquid  at  rest  is 
horizontal.  Assume  a  liquid  having  a  surface  which  is  not  hori- 
zontal, such  as  A  BCDE  (Fig.  ^  n >N 

4) .  A  plane  MN,  inclined  to 
the  horizontal,  may  be 
passed  through  any  liquid 
having  such  a  surface  in  such 
manner  that  a  portion  of  the 
liquid  BCD  lies  above  the  plane.  Since  the  liquid  is  at  rest,  BCD 
must  be  in  equilibrium,  but  the  vertical  force  of  gravity  would 
necessarily  have  a  component  along  the  inclined  plane  which 
could  be  resisted  only  by  a  tangential  stress.  As  liquids  are  in- 
capable of  resisting  tangential  stress  it  follows  that  the  free  surface 
must  be  horizontal. 

13.  Atmospheric  Pressure. — All  gases  possess  mass  and  con- 
sequently have  weight.     The  atmosphere,  being  a  fluid  composed 
of  a  mixture  of  gases,  exerts  a  pressure  on  every  surface  with 
which  it  comes  in  contact.     At  sea  level  under  normal  conditions 
atmospheric  pressure  amounts  to  2116  Ibs.  per  square  foot  or  about 
14.7  Ibs.  per  square  inch. 

VARIATION  IN  ATMOSPHERIC  PRESSURE  WITH  ALTITUDE 


FIG.  4. 


Altitude  above 
sea  level  in  feet 

Pressure  in 
pounds  per 
square  inch 

Altitude  above 
sea  level  in  feet 

Pressure  in 
pounds  per 
square  inch 

0 

14.7 

5,280 

12.0 

1000 

14.15 

6,000 

11.7 

2000 

13.6 

7,000 

11.3 

3000 

13.1 

8,000 

10.9 

4000 

12.6 

9,000 

10.5 

5000 

12.1 

10,000 

10.1 

10  PRINCIPLES  OF  HYDROSTATIC  PRESSURE 

The  intensity  of  atmospheric  pressure  decreases  with  the 
altitude.  Owing  to  compressibility,  the  density  of  air  also 
decreases  with  altitude,  and  therefore  the  intensity  of  pressure 
changes  less  rapidly  as  the  altitude  increases.  The  accompanying 
table  gives  approximate  values  of  the  atmospheric  pressure  corre- 
sponding to  different  elevations  above  sea  level. 

14.  Vacuum. — A  perfect  vacuum,  that  is,  a  space  in  which 
there  is  no  matter  either  in  the  solid,  liquid  or  gaseous  form,  has 
never  been  obtained.     It  is  not  difficult,  however,  to  obtain  a 
space  containing  a  minute  quantity  of  matter.     A  space  in  contact 
with  a  liquid,  if  it  contains  no  other  substance,  always  contains 
vapor  from  that  liquid.     In  a  perfect  vacuum  there  could  be  no 
pressure. 

In  practice,  the  word  "vacuum"  is  used  frequently  in  con- 
nection with  any  space  having  a  pressure  less  than  atmospheric 
pressure,  and  the  term  " amount  of  vacuum"  means  the  amount 
the  pressure  is  less  than  atmospheric  pressure.  The  amount 
of  vacuum  is  usually  expressed  in  inches  of  mercury  column 
or  in  pounds  per  square  inch  measured  from  atmospheric  pressure 
as  a  base.  For  example,  if  the  pressure  within  a  vessel  is  reduced 
to  12  Ibs.  per  square  inch,  which  is  equivalent  to  24.5  in.  of  mer- 
cury column,  there  is  said  to  be  a  vacuum  of  2.7  Ibs.  per  square 
inch  or  5.5  inches  of  mercury.  (Arts.  20  and  22.) 

15.  Absolute  and  Gage  Pressure. — The  intensity  of  pressure 
above  absolute  zero  is  called  absolute  pressure.     Obviously,   a 
negative  absolute  pressure  is  impossible. 

Usually  pressure  gages  are  designed  to  measure  the  intensities 
of  pressure,  above  or  below  atmospheric  pressure  as  a  base.  Pres- 
sures so  measured  are  called  relative  or  gage  pressures.  Negative 
gage  pressures  indicate  the  amount  of  vacuum,  and  at  sea  level 
pressures  as  low  as,  but  no  lower  than,  — 14.7  Ibs.  per  square  inch 
are  possible.  Absolute  pressure  is  always  equal  to  gage  pressure 
plus  atmospheric  pressure. 

Fig.  5  illustrates  a  gage  dial,  on  the  inner  circle  of  which  is 
shown  the  ordinary  gage  and  vacuum  scale.  The  outer  scale 
indicates  the  corresponding  absolute  pressures. 

16.  Intensity  of  Pressure  at  any  Point. — To  determine  the 
intensity  of  pressure  at  any  point  in  a  liquid  at  rest  or  the  variation 
in  pressure  in  such  a  liquid,  consider  any  two  points  such  as  1  and 
2  (Fig.  6)  whose   vertical  depths  below  the  free    surface  of  the 


INTENSITY  OF  PRESSURE  AT  ANY  POINT 


11 


liquid  are  hi  and  fa,  respectively.  Consider  that  these  points  lie 
in  the  ends  of  an  elementary  prism  of  the  liquid,  having  a  cross- 
sectional  area  dA  and  length  I.  Since  this  prism  is  at  rest,  all  of 
the  forces  acting  upon  it  must  be  in  equilibrium.  These  forces 
consist  of  the  fluid  pressure  on  the  sides  and  ends  of  the  prism 
and  the  force  of  gravity, 


FIG.  5.  —  Gage  dial. 

Let  X  and  Y,  the  coordinate  axes,  be  respectively  parallel 
with  and  perpendicular  to  the  axis  of  the  prism  which  makes  an 
angle  6  with  the  vertical.  Also  let  pi  and  p%  be  the  intensities  of 
pressure  at  points  1  and  2,  respectively,  and  w  be  the  unit  weight  of 
the  liquid. 

Considering  forces  acting  to  the  left  along  the  J^-axis  as 
negative  and  remembering  that  the  pressures  on  the  sides  of  the 
prism  are  normal  to  the  .XT-axis  and  therefore  have  no  X  com- 
ponents, the  following  equation  may  be  written: 


=  p  id  A  —  p2dA  —  wldA  cos  0=0. 
Since  I  cos  B  =  hi  —  fa,  this  reduces  to 


(3) 


From  this  equation  it  is  evident  that  in  any  liquid  the  difference 
in  pressure  between  any  two  points  is  the  product  of  the  unit 
weight  of  the  liquid  and  the  difference  in  elevation  of  the  points. 

If  hi  =  fa,  pi  must  equal  p2 ;  or,  in  other  words,  in  any  continuous 
homogeneous  body  of  liquid  at  rest,  the  intensities  of  pressure  at 
all  points  in  a  horizontal  plane  must  be  the  same.  Stated  con- 


12  PRINCIPLES  OF  HYDROSTATIC   PRESSURE 

versely,  in  any  homogeneous  liquid  at  rest  all  points  having  equal 
intensities  of  pressure  must  lie  in  a  horizontal  plane. 

If  in  equation  (3)  hz  is  made  equal  to  zero,  p2  is  the  intensity  of 
pressure  on  the  liquid  surface.  In  case  that  pressure  is  atmos- 
pheric, or  pa,  equation  (3)  becomes 

pi=whi+pa, (4) 

or,  in  general, 

p  =  wh+pa.       .     .     .     .     .     .     .     (5) 

In  this  equation  p  is  evidently  the  absolute  pressure  at  any  point 
in  the  liquid  at  a  depth  h  below  the  free  surface.  The  correspond- 
ing gage  pressure  is 

p  =  wh (6) 

In  the  use  of  the  above  equations  care  must  be  taken  to  express 
all  of  the  factors  involved  in  their  proper  units.  Unless  otherwise 
stated  p  will  always  be  understood  to  be  intensity  of  pressure  in 
pounds  per  square  foot,  w  will  be  the  weight  of  a  cubic  foot  of  the 
liquid  and  h  will  be  measured  in  feet. 

At  any  point  in  a  body  of  water  at  a  depth  h  below  the  free 
surface,  the  absolute  pressure  in  pounds  per  square  foot  is 

p  =  Q2. 4/1+2116.  . (7) 

The  relative  or  gage  pressure  in  pounds  per  square  foot  is 

p  =  62Ah (8) 

If,  however,  it  is  desired  to  express  the  pressure  in  pounds  per 
square  inch  it  is  necessary  only  to  divide  through  by  144.  Hence 
if  p'  is  used  to  express  absolute  pressure  in  pounds  per  square 
inch, 

j^G^      2116 
"      144      144     r  144 

=  .433/i+14.7, 

or,  expressed  as  gage  pressure  in  pounds  per  square  inch, 
p'=.  433/1. 


PRESSURE  HEAD  13 

17.  Pressure  Head.  —  Equation  (6)  may  be  written  in  the  form, 

2  =  h.  (9) 

W 
T) 

Here  h,  or  its  equivalent,  —  ,  represents  the  height  of  a  column  of 

liquid  of  unit  weight  w  that  will  produce  an  intensity  of  pressure, 
p.     It  is  therefore  called  pressure  head. 

In  considering  water  pressures  the  pressure  head,  h,  is  expressed 
in  feet  of  water  column  regardless  of  whether  it  is  obtained  by 
dividing  the  pressure  in  pounds  per  square  foot  by  62.4  or  by 
dividing  the  pressure  in  pounds  per  square  inch  by  0.433. 

18.  Transmission  of  Pressure.  —  Writing  equation  (3)  in  the 
form, 

h2),    ......     (10) 


it  is  evident  that  the  pressure  at  any  point,  such  as  point  1  (Fig.  6  ), 
in  a  liquid  at  rest  is  equal  to  the  pressure  at  any  other  point,  such 
as  point  2,  plus  the  pressure  produced  by  a  column  of  the  liquid 
whose  height,  h,  is  equal  to  the  difference  in  elevation  between  the 
two  points.  Any  change  in  the  intensity  of  pressure  at  point  2 
would  cause  an  equal  change  at  point  1.  In  other  words,  a  pres- 
sure applied  at  any  point  in  a  liquid  at  rest  is  transmitted  equally 
and  undiminished  to  every  other  point  in  the  liquid. 

This  principle  is  made  use  of  in  the  hydraulic  jack  by  means 
of  which  heavy  weights  are  lifted  by  the  application  of  relatively 
small  forces. 

Example.  —  In  Fig.  7  assume  that  the  piston  and  weight,  W, 
are  at  the  same  elevation,  the  face  of  the  piston  having  an  area  of 
2  sq.  in.  and  the  face  of  the 
weight  20  sq.  in.  What  weight 
W  can  be  lifted  by  a  force  P  of 
100  Ibs.  applied  at  the  end  of 
the  lever  as  shown  in  the 
figure? 

Since  atmospheric  pressure  FIG.  7.  —  Hydraulic  jack. 

is  acting   on   both  the  piston 
and  weight  its  resultant  effect  will  be  zero  and  it  may  therefore  be 


14 


PRINCIPLES  OF  HYDROSTATIC  PRESSURE 


neglected.     Taking  moments  about  0  the  force  F  on  the  piston  is 
§F  =  4X100 
F  =  600  Ibs. 

600 

——  =  300  Ibs.  per  square  inch. 
& 

which  is  the  intensity  of  pressure  on  the  face  of  the  piston,  and 
since  the  two  are  at  the  same  elevation  in  a  homogeneous  liquid 
at  rest  it  is  also  the  intensity  of  pressure  on  the  weight.  Therefore 


=  6000  Ibs. 

Evidently  this  is  the  value  of  W  for  equilibrium;  any  weight  less 
than  6000  pounds  could  be  lifted  by  the  force  of  100  Ibs. 

19.  Vapor  Pressure.  —  Whenever  the  free  surface  of  any  liquid 
is  exposed  to  the  atmosphere,  evaporation  is  continually  taking 
place.  If,  however,  the  surface  is  in  contact  with  an  enclosed 
space,  evaporation  takes  place  only  until  the  space  becomes 
saturated  with  vapor.  This  vapor  produces  a  pressure,  the 
amount  of  which  depends  only  upon  the  temperature  and  is 
entirely  independent  of  the  presence  or  absence  of  air  or  other 
gas  within  the  enclosed  space.  The  pressure  exerted  by  a  vapor 
within  a  closed  space  is  called  vapor  pressure. 

In  Fig.  8,  A  represents  a  tube  having  its  open  end  submerged 

in  water  and  a  stopcock  at  its 
upper  end.  Consider  the  air 
within  A  to  be  absolutely  dry 
at  the  time  the  stopcock  is 
closed.  At  the  instant  of 
closure  the  water  surfaces  in- 
side and  outside  the  tube  will 
stand  at  the  same  level. 
Evaporation  within  the  tube, 
however,  will  soon  saturate  the 
space  containing  air  and  create 
a  vapor  pressure,  pv,  which 
will  cause  a  depression  of  the 


Pa  ~ 


FIG.  8. 


water  surface  within  the  vessel  equal  to  — . 


THE   MERCURY  BAROMETER 


15 


In  the  same  figure,  B  represents  a  tube  closed  at  the  upper  end. 
Assume  a  perfect  vacuum  in  the  space  above  the  water  in  the  tube. 
If  this  condition  were  possible  the  water  level  in  B  would  stand 

at  an   elevation   —  above   the   water  surface   outside.      Vapor 

w 

pressure  within  the  vessel,  however,  causes  a  depression  —  exactly 

w 

equal  to  that  produced  within  A,  so  that  the  maximum  height  of 
water  column  possible  under  conditions  of  equilibrium  in  such  a 

tube  is  — — ~.     Vapor  pressures  increase  with  the  temperature, 
as  is  shown  in  the  following  table. 

VAPOR  PRESSURES  OF  WATER  IN  FEET  OF  WATER  COLUMN 


Tempera- 
ture, F. 

EL 

w 

Tempera- 
ture, F. 

Pv_ 
W 

Tempera- 
ture, F. 

EL 

w 

-20° 

0.02 

60° 

0.59 

140° 

6.63 

-10 

.03 

70 

0.83 

150 

8.54 

0 

.05 

80 

1.16 

160 

10.90 

10 

.08 

90 

1.59 

170 

13.78 

20 

.13 

100 

2.17 

180 

17.28 

30 

.19 

110 

2.91 

190 

21.49 

40 

.28 

120 

3.87 

200 

26.52 

50 

.41 

130 

5.09 

212 

33.84 

20.  The  Mercury  Barometer. — The  barometer  is  a  device  for 
measuring  intensities  of  pressure  exerted  by  the 
atmosphere.  In  1643  Torricelli  discovered  that 
if  a  tube  (Fig.  9)  over  30  in.  long  and  closed  at 
one  end,  is  rilled  with  mercury  and  then  made 
to  stand  vertically  with  the  open  end  submerged 
in  a  vessel  of  mercury,  the  column  in  the  tube 
will  stand  approximately  30  in.  above  the  sur- 
face of  the  mercury  in  the  vessel.  Such  a  device 
is  known  as  a  mercury  barometer.  Pascal  proved 
that  the  height  of  the  column  of  mercury  depended 
upon  the  atmospheric  pressure,  when  he  carried  a  FlG-  9- — Mercury 


barometer  to  a  higher  elevation  and  found  that 


barometer. 


the  height  of  the  column  decreased  as  the  altitude  increased. 


16 


PRINCIPLES  OF  HYDROSTATIC  PRESSURE 


Although  theoretically  water  or  any  other  liquid  may  be  used 
for  barometers,  two  difficulties  arise  in  using  water.  First,  the 
height  of  water  column  necessary  to  balance  an  atmospheric  pres- 
sure of  14.7  Ibs.  per  square  inch  is  about  34  feet  at  sea  level,  which 
height  is  too  great  for  convenient  use;  and,  second,  as  shown  in 
Art.  19,  water  vapor  collecting  in  the  upper  portion  of  the  tube 
creates  a  pressure  which  partially  balances  the  atmospheric  pres- 
sure, so  that  the  barometer  does  not  indicate  the  total  atmospheric 
pressure.  ^ 

Since  mercury  is  the  heaviest  known  liquid  and  has  a  very  low 
vapor  pressure  at  ordinary  air  temperatures  it  is  more  satisfactory 
for  use  in  barometers  than  any  other  liquid. 

21.  Piezometer  Tubes. — A  piezometer  tube  is  a  tube  tapped 
into  the  wall  of  a  vessel  or  pipe  for  the  purpose 
of  measuring  moderate  pressures.  Thus  the 
height  of  water  column  in  tube  a  (Fig.  10)  is 
a  measure  of  the  pressure  at  J.,  the  top  of  the 
pipe.  Similarly  the  pressure  at  the  elevation  B 
is  measured  by  the  height  of  water  column  in 
tube  6,  that  is,  p  =  wh.  Piezometer  tubes 
always  measure  gage  pressures  since  the  water 
surface  in  the  tube  is  subjected  to  atmospheric 
pressure.  Obviously,  the  level  to  which  water 
will  rise  in  a  tube  will  be  the  same  regardless 
of  whether  the  connection  is  made  in  the  side, 
bottom  or  cover  of  the  containing  vessel. 

Piezometer  tubes  are  also  used  to  measure  pressure  heads  in 
pipes  where  the  water  is  in  motion.  Such  tubes  should  enter  the 
pipe  in  a  direction  at  right  angles  to  the  direction  of  flow  and  the 
connecting  end  should  be  flush  with  the  inner  surface  of  the  pipe. 
If  these  precautions  are  not  observed,  the  height  of  water  column 
may  be  affected  by  the  velocity  of  the  water,  the  action  being 
similar  to  that  which  occurs  in  Pitot  tubes.  (See  Art.  48.) 

In  order  to  avoid  the  effects  of  capillary  action,  piezometer 
tubes  should  be  at  least  ^  in.  in  diameter. 

Pressures  less  than  atmospheric  pressure  may  be  measured  by 
either  of  the  methods  illustrated  in  Fig.  11  which  shows  a  pipe 
section  AB  in  which  the  water  is  flowing.  The  vertical  distance, 
ht  which  the  water  surface,  C,  in  the  open  tube  drops  below  A 
is  a  measure  of  the  pressure  below  atmospheric  pressure,  or,  in 


FIG.  10.  —  Cross- 
section  of  pipe 
with  piezometer 
tubes. 


MERCURY  GAGE 


17 


other  words,  it  is  a  measure  of  the  amount  of  vacuum  existing  at 
A.  This  is  true  since  the  pressure  at  Cf  is  atmospheric,  being  at 
the  same  level  as  C  in  a  homogeneous  liquid  at  rest  and  the 
pressure  at  A  must  be  less  than  at  C'  by  the  amount  wh.  (See 
Art.  17.) 

To  the  right  in  Fig.  11  is  shown  an  inverted  piezometer  tube 
with  the  lower  end  immersed  in  an  open  vessel  containing  water. 
Atmospheric  pressure  acting  on  the  water  surface  in  the  vessel 
forces  water  to  rise  in  the  tube  to  a  height  h  which  measures  the 
vacuum  existing  at  B.  Air  will  be  drawn  into  the  pipe  at  B 
until  the  intensity  of  pressure  on  the  water  surface  in  the  tube 
equals  the  pressure  at  B.  Neglecting  the  weight  of  the  air  in 
the  tube,  it  is  evident  that  pn  =  pD  =  pa-  wh.  Here  pB  is  expressed 
as  absolute  pressure  since  atmospheric  pressure  is  included  in 
the  equation, 


FIG.  11. 


FIG.  12. — Mercury  gage. 


22.  Mercury  Cage.  —  In  the  measurement  of  pressures  so  great 
that  the  length  of  tube  required  for  a  water  piezometer  would  be 
unwieldy,  the  mercury  U-tube,  illustrated  in  Fig.  12,  is  a  con- 
venient substitute. 

Water  under  pressure  fills  the  pipe,  or  vessel  at  A  and  the  tube 
down  to  the  level  D.  Mercury  fills  the  tube  from  D  around  to  B, 
above  which  level  the  tube  is  open  to  the  atmosphere. 

The  pressure  at  C  equals  the  pressure  at  B  which  is  atmos- 
pheric, or  pa,  plus  the  pressure  produced  by  the  mercury  column, 
h.  Hence, 

pc=pa+w'h.     ,,,,,,,     (11) 


If  pc  and  pa  are  expressed  in  pounds  per  square  foot  and  h 
in  feet,  w'  is  the  weight  of  a  cubic  foot  of  mercury,  or  13.6  (specific 


18 


PRINCIPLES  OF  HYDROSTATIC   PRESSURE 


gravity  of  mercury)  X  62.  4  =  848  Ibs.     If,  however,  pc  and  pa  are 
expressed  in  pounds  per  square  inch  and  h  in  feet, 

040 
™'  =         (or  13.  6X0.  433)  =5.  89. 


The  pressure  at  D  (Fig.  12)  is  the  same  as  at  C,  being  at  the 
same  level  in  a  homogeneous  liquid  at  rest.  The  pressure  at  A 
is  equal  to  that  at  C  minus  the  pressure  produced  by  the  water 
column  a,  or, 

PA  =  PC—WCL  ........     (12) 

Here  again  if  pA  and  pc  are  expressed  in  pounds  per  square 
foot  and  a  in  feet,  w  equals  62.4,  but  if  pA  and  pc  are  expressed 
in  pounds  per  square  inch  and  a  in  feet,  w  equals  0.433.  Com- 
bining equations  (11)  and  (12), 

pA  =  pa-\-w'h—wa.    .     .     .     (13) 

Here  PA  is  expressed  as  absolute  pressure, 
since  pa  (atmospheric  pressure)  enters  into  the 
equation. 

If  PA  is  less  than  atmospheric  pressure  by 
an  amount  greater  than  wa,  h  is  negative,  as  in 
Fig.  13,  and 

w'h—wa.      .     .      (14) 


FIG.  13. 


23.  The  Differential  Gage.— The  dif- 
ferential gage  as  the  name  indicates,  is 
used  only  for  measuring  differences  in  pres- 
sure. A  liquid  heavier  or  lighter  than  water 
is  used  in  the  gage,  depending  upon  whether 
the  differences  in  pressure  to  be  measured 
are  great  or  small. 

In  Fig.  14  is  shown  the  form  of  dif- 
ferential gage  usually  employed  for  meas- 
uring large  differences  in  pressure.  M  and 
N  are  two  pipes  containing  water  under 
different  pressures  which  may  be  either 
greater  or  less  than  atmospheric  pressure. 
The  two  pipes  are  connected  by  a  bent 
tube,  of  which  the  portion  BCD  is  filled  with  mercury,  while  all 
of  the  remaining  space  is  filled  with  water. 


FIG.  14.— Differential 
mercury  gage. 


THE  DIFFERENTIAL  GAGE 


19 


If  M  and  N  were  at  the  same  elevation  the  difference  in  pres- 
sure in  the  two  pipes  would  be  measured  by  the  pressure  due  to  the 
mercury  column  DC  minus  that  due  to  the  water  column  EF 
or  would  equal  w'h—wh. 

If  the  two  pipes  are  at  different  elevations,  to  the  above 
difference  in  pressure  must  be  added  or  subtracted  the  intensity 
of  pressure  produced  by  the  water  column  whose  height  is  equal 
to  the  difference  in  elevation  of  the  pipes.  Proof  of  these  state- 
ments follows. 

The  pressure  at  A  equals  that  at  M  minus  the  pressure  produced 
by  the  water  column  whose  height  is  a.  Evidently  the  pressure 
at  B  is  the  same  as  at  A,  being  at  the  same  elevation  in  a  homo- 
geneous liquid  at  rest.  For  the  same  reason  the  pressures  at 
B  and  C  are  also  equal.  Hence, 

pc=pB  =  pA  =  pM—wa (15) 

The  pressures  at  C  and  F  are  not  equal,  since  these  points  are 
not  connected  by  a  homogeneous  liquid. 

The  pressure  at  D  is  equal  to  that  at  C  minus  the  pressure 
produced  by  the  mercury  column  h  and  is 


=  pc—w'h. 


(16) 


and 


Combining  these  equations 


or 


Obviously,  the  greater  the  difference  between 
wf  and  w  the  greater  the  difference  in  pressure 
that  can  be  measured  for  any  given  value 
of  h. 

Fig.  15  illustrates  a  type  of  differential 
gage  used  when  the  difference  in  pressures  to 
be  measured  is  small.  Usually  a  liquid,  such 
as  a  light  oil,  whose  specific  gravity  is  slightly 
less  than  unity  is  used  in  the  upper  portion  of  the  inverted  U-tube, 
AC,  the  remainder  of  the  tube  being  filled  with  water. 


FIG.  15. — Differential 
oil  gage. 


20 


PRINCIPLES  OF  HYDROSTATIC   PRESSURE 


As  with  the  mercury  gage,  if  M  and  N  are  at  the  same  elevation 
the  difference  in  pressure  will  be  equal  to  that  produced  by  the 
oil  column  AB  and  the  water  column  CD.  If  M  and  N  are  at 
different  levels,  then 


or 


....     (19) 
....     (20) 


the  equations  being  the  same  as  were  obtained  for  the  mercury 
differential  gage. 

The  use  of  a  liquid  whose  unit  weight  w'  is  very  nearly  the 
same  as  that  of  water  makes  a  very  sensitive  measuring  device. 
With  such  a  device  small  differences  in  pressure  will  produce 
relatively  large  values  of  h. 

The  value  of  h  produced  by  any  particular  difference  in  pressure 
is  independent  of  the  relative  cross-sectional  areas  of  the  columns 
AB  and  CD.  This  is  evident  since  it  is  the  difference  in  intensities 
of  pressure  that  is  measured  and  not  difference  in  total  pressures. 

24.  Suction  Pumps  and  Siphons. — Suction  pumps  depend  upon 
atmospheric  pressure  for  their  operation.  The  plunger  creates  a 
partial  vacuum  in  the  pump  stock,  and  atmospheric  pressure  acting 
upon  the  outer  water  surface  causes  water  to  rise  within  the 
pump. 

The  operation  of  siphons  is  also  produced  by  atmospheric 

pressure.  li\  Fig.  16  the  two 
vessels,  A  and  B,  are  connected 
by  a  tube.  As  long  as  the  tube 
is  filled  with  air  there  is  no 
tendency  for  water  to  flow.  If, 
however,  air  is  exhausted  from 
the  tube  at  C,  atmospheric 
pressure  will  cause  water  to  rise 
in  each  leg  of  the  tube  an  equal 
height  above  the  water  sur- 


FIG.  16. — Siphon. 


faces.  When  water  has  been  drawn  up  a  distance  d,  equal  to  the 
height  of  the  summit  above  the  water  surface  in  the  upper  vessel 
flow  from  A  to  B  will  begin.  If  the  velocity  of  the  water  is  high 
enough,  any  air  entrapped  in  the  tube  will  be  carried  out 
by  the  moving  water,  and  the  tube  will  flow  full.  If  the  sum- 


PROBLEMS  21 

mit  of  the  siphon  is  a  distance  greater  than  — — —  above  the 

10 

water  surface  in  the  higher  vessel,  siphon  action  is  impossible. 

It  is  not  necessary  that  the  discharge  end  of  the  tube  be  sub- 
merged to  induce  siphon  action.  If  flow  is  started  by  suction  at 
the  free  end  of  the  tube  or  by  other  means,  it  will  continue  as  long 
as  the  discharge  end  of  the  tube  is  lower  than  the  water  surface 
in  the  vessel  or  until  the  vacuum  in  the  siphon  is  broken. 

PROBLEMS 

1.  Determine  the  intensity  of  pressure  on  the  face  of  a.dam  at  a  point 
40  ft.  below  the  water  surface.  *#}t*}^ 

(a)  Expressed  in  pounds  per  square  foot  gage  pressure. 
(6)  Expressed  in  pounds  per  square  inch  gage  pressure. 

(c)  Expressed  in  pounds  per  square  foot  absolute  pressure. 

(d)  Expressed  in  pounds  per  square  inch  absolute  pressure. 

2.  Determine  the  intensity  of  pressure  in  a  vessel  of  mercury  (sp.  gr.  =  13.6) 
at  a  point  8  in.  below  the  surface,  expressing  the  answer  in  the  same  units  as 
in  Problem  1. 

3.  A  vertical  pipe,  100  ft.  long  and  1  in.  in  diameter,  has  its  lower  end  open 
and  flush  with  the  inner  surface  of  the  cover  of  a  box,  2  ft.  square  and  6  in. 
high.     The  bottom  of  the  box  is  horizontal.     Neglecting  the  weight  of  the 
pipe  and  box,  both  of  which  are  filled  with  water,  determine: 

(a)  The  total  hydrostatic  pressure  on  the  bottom  of  the  box. 
(6).  The  total  pressure  exerted  on  the  floor  on  which  the  box  rests. 

4.  At  what  height  will  water  stand  in  a  water  barometer  at  an  altitude 
of  5000  ft.  above  sea  level  if  the  temperature  of  the  water  is  70°  F.?     Under 
similar  conditions  what  would  be  the   reading  of  a   mercury   barometer, 
neglecting  the  vapor  pressure  of  mercury? 

5.  What  are  the  absolute  and  gage  pressures  in  pounds  per  square  inch 
existing  in  the  upper  end  of  the  water  barometer  under  the  conditions  of 
Problem  4? 

6.  What  height  of  mercury  column  will  cause  an  intensity  of  pressure  of 
100  Ibs.  per  square  inch?     What  is  the  equivalent  height  of  water  column? 

7.  A  pipe  1  in.  in  diameter  is  connected  with  a  cylinder  24  in.  in  diameter, 
each  being  horizontal  and  fitted  with  pistons.     The  space  between  the  pistons 
is  filled  with  water.     Neglecting  friction,  what  force  will  have  to  be  applied 
to  the  larger  piston  to  balance  a  force  of  20  Ibs.  applied  to  the  smaller  piston? 

8.  In  Problem  7,  one  leg  of  a  mercury  U-tube  is  connected  with  the 
smaller  cylinder.     The  mercury  in  this  leg  stands  30  in.  below  the  center  of 
the  pipe,  the  intervening  space  being  filled  with  water.     What  is  the  height 
of  mercury  in  the  other  leg,  the  end  of  which  is  open  to  the  air  ? 

9.  A  U-tube  with  both  ends  open  to  the  atmosphere  contains  mercury 
in  the  lower  portion.     In  one  leg,  water  stands  30  in.  above  the  surface  of  the 
mercury;   in  the  other  leg,  oil  (sp.  gr.=0.80)  stands  18  in.  above  the  surface 


22  PRINCIPLES  OF  HYDROSTATIC  PRESSURE 

of  the  mercury.    What  is  the  difference  in  elevation  between  the  surf: 
the  oil  and  water  columns? 

10.  Referring  to  Fig.  15,  page  19,  if  the  pressure  at  M  is  20  Ibs.  per 
inch,  what  is  the  corresponding  pressure  at  N  if  a  =  1  ft.,  b  =  4  ft.  and  h  •- 
(Sp.gr.  of  oil  =  0.80.) 

11.  In  Fig.  15,  page  19,  determine  the  value  of  h,  if  a  =  1  ft.,  b  =  4  ft.  a 
pressure  at  N  is  1.4  Ibs.  per  square  inch  greater  than  at  M.     (Sp.  gr 
0.80.) 

12.  A  vertical  tube  10  ft.  long,  with  its  upper  end  closed  and  low 
open,  has  its  lower  end  submerged  4  ft.  in  a  tank  of  water.    Neglecting 
pressure,  how  much  will  the  water  level  in  the  tube  be  below  the  level 
tank? 

13.  In  Fig.  7,  page  13,  if  the  diameters  of  the  two  cylinders  are  3  : 
24  in.  and  the  face  of  the  smaller  piston  is  20  ft.  above  the  face  of  the 
W,  what  force  P  is  required  to  maintain  equilibrium  if  TF  =  8000  Ibs.? 

14.  Referring  to  Fig.  12,  page  17,  if  A  =  20  in.  and  a  =  12  in.,  what 
absolute  pressure  in  pounds  per  square  inch  at  A1    What  is  the  gage  pn 

15.  In  Problem  14,  if  the  surface  of  the  mercury  column  in  each  leg 
U-tube  stands  at  the  same  elevation  as  A  when  the  pressure  at  A  is 
pheric,  determine  the  values  of  a  and  h  when  the  gage  pressure  at  A  is 
per  square  inch,  the  diameter  of  the  tube  being  the  same  throughout. 

16.  Referring  to  Fig.  13,  page  18,  determine  the  absolute  press 
pounds  per  square  inch  at  A  when  a =8  in.  and  h  =  10  in.     What  is  the 
spending  gage  pressure? 

17.  In  Fig.  14,  page  18,  let  c  =  6  ft.,  and  assume  that  A  =  0  wh< 
pressure  at  M  is  atmospheric.  If  the  pressure  at  N  remains  constant,  det 
the  value  of  h  when  the  gage  pressure  at  If  is  increased  to  8  Ibs.  per 
inch. 

18    In  Fig.  14,  page  18,  if  a  =  24  in.  and  c=6  ft.,  what  is  the  vali 
when  the  pressure  at  M  is  10  Ibs.  per  square  inch  greater  than  at  N. 

19.  A  and  B  are,  respectively,  the  closed  and  open  ends  of  a  I 
both  being  at  the  same  elevation.     For  a  distance  of  18  in.  below 
tube  is  filled  with  oil  (sp.  gr.  =0.8);  for  a  distance  of  3  ft.  below  B,  th 
is  filled  with  water,  on  the  surface  of  which  atmospheric  pressure  is 
The  remainder  of  the  tube  is  filled  with  mercury.     What  is  the  al 
pressure  at  A  expressed  in  pounds  per  square  inch? 

20.  In  Problem  19,  if  B  were  closed  and  A  were  open  to  the  atmos 
what  would  be  the  gage  pressure  at  B,  expressed  in  pounds  per  square 


CHAPTER  III 
PRESSURE   ON   SURFACES 

25.  Total  Pressure  on  Plane  Areas. — The  total  pressure  on 
any  plane  surface  is  equal  to  the  product  of  its  area  and  the 
intensity  of  pressure  at  its  center  of  gravity.  This  may  be  proved 
as  follows: 

Fig.  17  shows  projections  on  two  vertical  planes  normal  to 


(a) 


(*) 


FIG.  17. 


each  other,  of  any  plane  surface,  MN,  subjected  to  the  full  static 
pressure  of  a  liquid  with  a  free  surface.  Projection  (6)  is  on  a 
plane  at  right  angles  to  MN.  The  surface  MN  makes  any  angle, 
6,  with  the  horizontal,  and,  extended  upward,  the  plane  of  this 
surface  intersects  the  surface  of  the  liquid  in  the  line  00,  shown 
as  the  point  0  in  (b). 

Consider  the  surface  MN  to  be  made  up  of  an  infinite  number 
of  horizontal  strips  each  having  a  width  dy  so  small  that  the 


24  PRESSURE  ON  SURFACES 

intensity  of  pressure  on  it  may  be  considered  constant.  The  area 
of  any  strip  whose  length  is  x,  is 

dA  =  xdy. 

The  liquid  having  a  unit  weight  of  w,  the  intensity  of  pressure  on 
any  strip  at  a  depth  h  below  the  surface  and  at  a  distance  y  from  the 
line  00  is 

p  =  wh  =  wy  sin  6. 

The  total  pressure  on  the  strip  is 

dP  =  wy  sin  6  dA 
and  the  total  pressure  on  MN  is 

(1) 
From  the  definition  of  center  of  gravity, 

JydA=Ay', ,     .     ,     (2) 

where  y'  is  the  distance  from  the  line  00  to  the  center  of  gravity 
of  A.  Hence, 

P  =  wsm  S  Ay' (3) 

Since  the  vertical  depth  of  the  center  of  gravity  below  the  water 
surface  is 

\    h'  =  y'  sin0,    .     .     .^    ....     (4) 
it  follows  that 

P  =  wh'A,    '  .   Jj.<  X     ...     (5) 

where  wh'  represents  the  intensity  of  pressure  at  the  center  of 
gravity  of  A. 

26.  Center  of  Pressure  on  Plane  Areas. — Any  plane  surface 
subjected  to  hydrostatic  pressure  is  acted  upon  by  an  infinite 
number  of  parallel  faNs%;vvhose  magnitudes  vary  with  the  depth, 
below  the  free  surface,  of  the  various  infinitesimal  areas  on  which 
the  respective  forces  act.  Since  these  forces  are  parallel  they  may 
be  replaced  by  a  single  resultant  force  P.  The  point  on  the 
surface  at  which  this  resultant  force  acts  is  called  the  center  of 
pressure.  In  other  words,  if  the  total  hydrostatic  pressure  on  any 
area  were  applied  at  the  center  of  pressure  the  same  effect  would  be 


CENTER  OF  PRESSURE  ON  PLANE  AREAS  25 

produced  on  the  area  as  is  produced  by  the  variable  pressures 
distributed  over  the  area. 

The  position  of  the  horizontal  line  containing  the  center  of 
pressure  of  a  plane  surface  subjected  to  hydrostatic  pressure  may 
be  determined  by  taking  moments  of  all  the  forces  acting  on  the 
area  about  some  horizontal  axis  in  the  plane  of  the  surface.  For 
the  case  described  in  the  preceding  article  and  illustrated  in 
Fig  17,  the  line  00  may  be  taken  as  the  axis  of  moments  for  the 
surface  MN.  Designating  by  y  the  distance  to  the  center  of 
pressure  from  the  axis  of  moments,  it  follows  from  the  defini- 
tion of  center  of  pressure  that, 


........     (6) 

or 


fydP 

J— 


y=—p-  .........   (7) 

It  was  shown  in  Art.  25  that 

dP  =  wy  sin  0  dA 
and 

P  =  wsin  BAy'  .....  !  .     .     (3) 

Substituting  these  values,  equation  (7)  becomes 

w  sin  6  I  y2dA      \  y2dA 
V  =    w  sine  Ay'  ~~     Ay'    '     .....     (8) 

in  whichjy2dA  is  the  moment  of  inertia,  /o,  of  MN  with  respect 
to  the  axis  OO,  and  Ay'  is  the  statical  moment,  S,  of  MN  with 
respect  to  the  same  axis. 
Therefore, 


Since  the  moment  of  inertia  of  an  area  about  any  axis  equals 
the  moment  of  inertia  of  the  area  about  a  parallel  axis  through  its 
center  of  gravity  plus  the  product  of  the  area  and  the  square  of  the 
distance  between  the  two  axes,  equation  (9)  may  be  written, 


26 


or 


PRESSURE  ON   SURFACES 

y=y'+- 


y 


(ii) 


where  k  is  the  radius  of  gyration  of  the  area. 

The  above  discussion  refers  only  to  the  determination  of 'the 
position  of  the  horizontal  line  which  contains  the  center  of  pressure 
—that  is,  y  gives  only  the  distance  from  the  horizontal  axis  of 
moments  to  the  center  of  pressure.  For  any  figure  such  that  the 
locus  of  the  midpoints  of  the  horizontal  strips  is  a  straight  line, 
as,  for  instance,  a  triangle  or  trapezoid  with  base  horizontal,  the 
center  of  pressure  falls  on  that  straight  line.  It  is  with  such 
figures  that  the  engineer  is  usually  concerned.  For  other  figures, 
the  horizontal  location  of  the  center  of  pressure  may  be  found  in 


T 


t/V 


FIG.  18. 

a  manner  similar  to  that  described  above  by  taking  moments 
about  an  axis,  within  the  plane  of  the  surface,  at  right  angles  to  the 
horizontal  axis  of  moments. 

Examples  l :    (a)  Find  the  center  of  pressure  on  the  vertical 
triangular  gate  shown  in  Fig.  18. 

1  It  is  apparent  that  the  solution  of  any  problem  involving  the  location  of  the 
center  of  pressure,  for  an  area  whose  radius  of  gyration  is  known  or  can  be 
readily  found,  may  be  accomplished  by  the  simple  substitution  of  values  for 
k  and  y'  in  equation  (11).  This  involves  a  mere  mathematical  process  with 
no  necessity  on  the  part  of  the  student  for  either  thought  or  understanding 
of  fundamental  principles.  It  is  therefore  recommended  that  the  beginner 
solve  all  such  problems  by  the  use  of  equation  (6)  which  is  really  nothing 
more  than  a  formulated  expression  of  the  definition  of  center  of  pressure. 
The  examples  given  are  solved  by  this  method. 


CENTER  OF  PRESSURE  ON  PLANE  AREAS 


27 


Using  the  equation 


Py  =  j'ydP 


(6) 


it  will  be  necessary  to  express  dP  in  terms  of  y. 

In  this  case  it  will  be  convenient  to  take  moments  about  0-0, 
a  horizontal  line  through  the  vertex  of  the  gate,  lying  in  the  plane 
of  the  gate.  Moments  could,  however,  be  taken  about  any  other 
horizontal  axis  lying  within  this  plane. 

The  total  pressure  dP  on  any  thin  horizontal  strip  at  a  distance 
y  from  the  axis  of  moments  equals  the  intensity  of  pressure,  wh, 
times  the  area  dA,  or 

dP  =  whdA. 


Since  w  =  62A,  h  =  5+y  and  dA=x  dy 

dP=62A(5+y)xdy. 

Since  x  varies  with  y  it  must  be  expressed  in  terms  of  y  before 
integrating. 

From  similar  triangles, 


Substituting, 


x    y  4 

4  =  3    °r    x  =  &- 


and 

also 
therefore 


(5y+y2)dy 


fydP 
J 


wh'A  =  62.  4X7X6, 


« 
y 


62.4X42 


=  A(*X27+iX81) 

=  2^  ft.  below  the  vertex, 

or  ^  ft.  below  the  center  of  gravity  of  the  gate. 

The  horizontal  location  of  the  center  of  pressure  in  this  case 
is  on  the  median  connecting  the  vertex  with  the  base. 


28 


PRESSURE  ON  SURFACES 


(6)  Find  the  center  of  pressure  on  the  inclined  rectangular 
gate  shown  in  Fig.  19. 

Taking  moments  again  about  the  top  of  the  gate  as  an  axis, 


where 


and 


j  y  dP  —  \  ywh  dA, 


dA=My 


5' 


FIG.  19. 
Since  P  =  whf A  =62.4(5+2  cos  30°; 


62.4X6  J    Uy+ 


62.4X24(5+V3) 
1 


5  2     V3     I4 


I4 
J0 


^58.47 
26.93 

=  2.  17  ft.  below  AB, 

measured  along  the  plane  of  the  gate.     The  horizontal  location 
of  the  center  of  pressure  is  3  ft.  from  either  end  of  the  gate. 


GRAPHICAL   METHOD  OF  LOCATION  29 

(c)  In  example  (6)  what  force  F  applied  normally  to  the  gate 
at  its  lower  edge  will  be  required  to  open  it? 

Knowing  the  total  pressure,  P,  on  the  gate  and  the  location 
of  the  center  of  pressure,  by  taking  moments  about  the  upper 
edge  which  is  the  center  of  rotation, 

4F  =  2.17P 

. 4X24(5+ \/3) 


=  5470lbs. 

The  value  of  F  can  also  be  found  directly  without  determining 
either  the  location  of  the  center  of  pressure  or  the  total  pressure. 
Taking  moments  about  the  top  of  the  gate, 

4F=  CydP 

JQ 

r 

=  I    ywh  dA 

=  62.4X6  J     (5y+-^y2}dy 

=  62.4X6  |i/2+-—?/3 

V3 


=  62.4X6^X16+X64) 


=  21890 
F  =  5470lbs. 

If  this  force  were  applied  at  the  bottom  of  the  gate,  the  gate 
would  be  in  equilibrium  and  there  would  be  no  reaction  on  the 
supports  along  the  lower  edge  or  sides  of  the  gate.  Any  force 
greater  than  5470  Ibs.  would  open  the  gate. 

27.  Graphical  Method  of  Location  of  Center  of  Pressure.— 
Semigraphic  methods  may  be  used  advantageously  in  locating  the 
center  of  pressure  on  any  plane  area  whose  width  is  constant. 


30  PRESSURE  ON   SURFACES 

The  rectangular  surface,  A  BCD,  illustrated  in  Fig.  19  is  shown  in 
perspective  in  Fig.  20a.  BC  (Fig.  206)  represents  the  projection 
of  the  rectangle  on  a  vertical  plane  perpendicular  to  the  plane  of 
the  surface.  The  vertical  depths  below  the  water  surface  of  the 
top  and  bottom  of  the  rectangle  are,  respectively,  hi  and  h%.  The 
intensity  of  pressure,  whi,  on  the  top  of  the  rectangle  is  represented 
by  the  equal  ordinates  A  A'  and  BB'  (Fig.  20a),  and  on  the  bottom 
of  the  rectangle  the  equal  ordinates  CC'  and  DD'  represent  the 
intensity  of  pressure  whz. 

BB1  and  CC'  (Fig.  206)  represent,  to  a  reduced  scale,  the  inten- 
sities of  pressure  at  B  and  C.    BB"  and  CC"  are  laid  off  equal 


(a)  (6) 

FIG.  20. 

to  6BB;  and  6CC',  respectively,  and  therefore  represent  the  areas 
ABB' A'  and  DCC'D'.  The  total  pressure  acting  on  the  surface 
ABCD  is  therefore  represented  by  the  area  of  the  pressure  diagram 
BCC"B"  as  it  is  similarly  represented  by  the  pressure  volume 
ABCDA'B'C'D'.  Also  the  resultant  pressure  on  the  surface  acts 
through  the  center  of  gravity  of  the  pressure  area  BCC"B"  just 
as  it  acts  through  the  center  of  gravity  of  the  pressure  volume 
ABCDA'B'C'D'. 

The  trapezoid  BCC"B"  may  be  divided  into  the  rectangle 
BB"EC  and  the  triangle  B"C"E,  the  locations  of  whose  centers  of 
gravity  are  known.  By  taking  moments  of  each  of  these  pressure 
areas  about  C"C  and  dividing  the  sum  of  these  moments  by  the 


POSITION  OF  CENTER  OF  PRESSURE  31 

area  of  the  trapezoid,  the  distance  of  the  center  of  pressure  from 
C  is  determined.     Thus 


and 

CC" 
therefore 

C"# 

Taking  moments  about  C"C, 
-= 


4Xjtu>X5X6+w(5+2V3)X6] 

=  1.83  ft. 

Example  (c),  page  29,  can  also  be  solved  by  taking  moments 
about  BB"  as  follows: 


4X5X6X2+fX62.  4X2V3X6XI 
from  which 

F  =  54701bs. 

For  areas  having  a  variable  width,  OB"C"  is  not  a  straight 
line  and  the  center  of  gravity  of  the  pressure  area  is  not  so  easily 
located.  For  such  areas  it  will  probably  be  easier  to  use  the 
analytical  method  described  in  Art.  26. 

28.  Position  of  Center  of  Pressure  with  Respect  to  Center  of 
Gravity.  —  If  the  intensity  of  pressure  varies  over  any  surface,  the 
center  of  pressure  is  below  the  center  of  gravity.  Consider  the 
equation  (see  Art.  26), 

£2 

~=y'+-,  .........  (ii) 


k2 
Since  —  must  always  be  positive,  y  must  be  greater  than  yf. 

This  may  also  be  seen  from  Fig.  20.  The  center  of  pressure 
on  A  BCD  is  the  normal  projection  on  that  plane  of  the  center  of 
gravity  of  the  pressure  volume  ABCDA'B'C'D'.  Evidently  this 
projection  must  fall  below  the  center  of  gravity  of  A  BCD  since 
it  would  fall  at  the  center  of  gravity  if  the  intensity  of  pressure 
on  the  surface  were  uniform,  in  which  case  the  pressure  volume 
would  be  ABCDA'B'EF. 


32 


PRESSURE  ON  SURFACES 


— N* 


It  also  appears  from  the  above  discussion  and  from  a  study  of 
Fig.  20  that  for  any  area  the  greater  its  depth  below  the  surface 
of  the  liquid  the  more  nearly  will  the  center  of  pressure  approach 
the  center  of  gravity.  The  two  coincide  at  an  infinite  depth. 

In  two  cases  the  intensity  of  pressure  is  constant  over  the  area 
and  hence  the  center  of  pressure  coincides  with  the  center  of 
gravity: 

(a)  When  the  surface  is  horizontal. 

(6)  When  both  sides  of  the  area  are  completely  submerged  in 
liquids  of  the  same  density.  As  an  illustration  consider  the  gate 
AB  (Fig.  21).  Water  stands  hi  feet  above  the  top  of  the  gate  on 
one  side  and  h?,  feet  above  it  on  the  other  side.  The  distribution 

of  pressure  on  the  left  is 
represented  by  the  trapezoid 
ABMN  and  on  the  right  by 
the  trapezoid  AHKB.  The 
triangle  GED  is  similar  to 
CFG  and  equal  to  CE'P  by 
construction.  The  trapezoid 
of  pressure  AHKB  is  therefore 
balanced  by  the  trapezoid 
ON  ML.  The  resultant  in- 
tensity of  pressure  on  the 
7  surface  is  therefore  constant, 
as  represented  by  the  rect- 
angle OABL,  and  the  center 

of  pressure  must  coincide  with  the  center  of  gravity  of  the 
surface.  This  is  true  regardless  of  the  shape  of  the  surface. 
The  resultant  intensity  of  pressure  equals  wh}  where  h  is  the 
difference  in  elevation  of  water  surfaces. 

In  this  latter  case  it  should  be  observed  that  it  is  the  center 
of  the  resultant  pressure  that  coincides  with  the  center  of  gravity 
of  the  gate,  since  the  center  of  gravity  of  either  of  the  trap- 
ezoidal areas  of  pressure,  considered  alone,  falls  below  the  center 
of  gravity  of  the  gate. 

29.  Horizontal  and  Vertical  Components  of  Pressure  on  any 
Surface. — It  may  be  convenient  to  deal  with  the  horizontal  and 
vertical  components  of  the  pressure  acting  on  a  surface  rather 
than  with  the  resultant  pressure. 

Consider,  for  example,  the  water  pressure  acting  on  the  curved 


7 


FIG.  21. 


HORIZONTAL  AND   VERTICAL  COMPONENTS 


33 


The  dam  may 


FIG.  22. 


face,  AB,  of  the  dam  shown  in  section  in  Fig.  22. 
have  any  length  normal  to  the 
plane  of  the  paper.  Choosing 
the  coordinate  axes  as  shown, 
let  BF  represent  the  trace  of 
a  vertical  plane  normal  to  the 
XY  plane.  Consider  the  equi- 
librium of  the  volume  of  liquid 
whose  cross-section,  as  shown 
in  the  figure,  is  ABF  and 
whose  ends  are  parallel  with 
the  XY  plane  and  separated 
by  a  distance  equal  to  the 
length  of  the  dam.  Since  this 
volume  of  liquid  is  assumed  to  be  in  equilibrium,  2X  =  0  and 
S7  =  0. 

The  only  forces  that  have  any  components  parallel  with  the 
X-axis  are  the  X-components  of  the  normal  pressures  acting  on  the 
surface  A B  and  the  normal  pressure  on  the  vertical  plan  BF. 
Since  BF  is  the  projection  of  the  face  A  B  on  a  vertical  plane  nor- 
mal to  the  X-axis,  it  follows  that  the  resultant  of  the  X-com- 
ponents  of  the  pressures  on  A  B,  or  Px,  equals  the  normal  pressure 
on  the  projection  of  A  B  on  a  vertical  plane  normal  to  the  X-axis. 
As  the  demonstration  holds  true  independently  of  the  manner  in 
which  the  X-axis  is  chosen,  it  may  be  stated  in  general  that  the 
component,  along  any  horizontal  axis,  of  the  pressure  on  any  area 
is  equal  to  the  normal  pressure  on  that  vertical  projection  of  the 
area  which  is  normal  to  the  chosen  axis. 

In  a  similar  manner  consider  the  vertical  forces  acting  on  the 
volume  of  liquid  whose  cross-section  is  ABF  (Fig.  22).  The  only 
vertical  forces  are  the  force  of  gravity,  or  the  weight  of  the  liquid, 
and  the  vertical  components  of  the  pressures  on  the  surface  AB, 
which  forces  must  therefore  be  equal  in  magnitude.  In  other 
words,  the  vertical  component  of  the  pressure  on  any  surface  is 
equal  to  the  weight  of  that  volume  of  the  liquid  extending  verti- 
cally from  the  surface  to  the  free  surface  of  the  liquid.  If  the  pres- 
sure were  acting  upward  on  the  surface,  its  magnitude,  as  will  be 
shown  later  (Art.  30),  would  be  equal  to  the  weight  of  that  volume 
of  the  liquid  that  would  extend  from  the  surface  to  the  free  surface 
of  the  liquid.  The  pressures  considered  in  this  article  are  relative 


34 


PRESSURE  ON   SURFACES 


pressures,  since  obviously  atmospheric  pressure  is  acting  on  both 
sides  of  the  dam  and  the  resultant  effect  is  zero. 

Examples:  (a)  What  will 
be  the  resultant  pressure  on 
the  base  BC  of  the  masonry 
dam  subjected  to  water  pres- 
sure, as  shown  in  Fig.  23,  and 
where  will  this  resultant  in- 
tersect the  base? 

The  following  numerical 
values  are  given.  Area  of 
section  ABCD  =  QQO  sq.  ft. 
Area  of  water  section  OAB 
=  200  sq.  ft.  Weight  of  ma- 
sonry =  150  Ibs.  per  cubic  foot. 
Linear  dimensions  are  shown 
in  figure. 

A  section  of  dam  1  ft.  long 
will  be  considered  to  be  in  equi- 
FIG.  23.  librium  under  the  action  of  the 

following  forces: 

TF  =  the  total  weight  of  the  section,  acting  through  the  center  of 

gravity  of  the  cross-section  A  BCD; 

P  =  resultant  hydrostatic  pressure  acting  on  the  face  AB; 
R  =  reaction  between  the  earth  and  the  base,  BC,  of  the  dam. 

This  reaction  must  necessarily  be  equal,  opposite  and  colinear 
with  the  resultant  of  W  and  P. 

Since  the  dam  is  in  equilibrium  when  subjected  to  the  above 
forces,  the  fundamental  principles  of  equilibrium  may  be  applied — 
that  is, 

2X  =  0,  27=0    and     2M  =  0.  > 


For  2X  =  0;  RX  =  PX  and  for  the  data  given 

Px  =  62 . 4 X4£ X45  =  63,200  Ibs.      \ 

For  27=0;  RV  =  PV+W  and  for  the  data  given 
Pv  =  200  X  62 . 4  =  12,480  Ibs.  • 
IF  =  600X150  =  90,000  Ibs. 
Rv=  12,480+90,000  =  102,480  Ibs. 


^ 


HORIZONTAL  AND   VERTICAL  COMPONENTS  35 

The  total  resultant  pressure  on  the  base  is 


63,2002+  102,4802  =  122,500  Ibs, 

For  2  Jlf  =  0,  taking  moments  about  an  axis  through  C,  normal  to 
the  plane  of  the  section,  X  being  the  distance  from  C  at  which  the 
resultant,  R,  intersects  the  base  of  the  dam, 


x-  25Py-  18W  =  0. 
Substituting  the  numerical  values  of  Rv,  Px,  W  and  Py 

102,480X+63,200X-V-~  12,480X25-90,000X18  = 
and  reducing 


Thus  the  resultant  pressure  of  122,500  Ibs.  per  linear  foot  of  dam 
intersects  the  base  8.6  ft.  from  the  toe  of  the  dam. 

(6)  Determine  the  tensile  stress  in  the  walls  of  a  24-in.  pipe 
carrying  water  under  a  head  of  100  ft. 

In  a  case  like  this  where  the  head  is  relatively  large  compared 
to  the  diameter  of  the  pipe,  it  is  customary  to  consider  that  the 
intensity  of  pressure  is  uniform  throughout  the  pipe. 

A  cross-section  of  the  pipe  is  shown  in  Fig.  24.  Consider  a 
semicircular  segment,  AB,  of  unit  length,  held  in  equilibrium  by 
the  two  forces  T.  Evidently 

T  is  the  tensile  stress  in  the   —  i  ---  >TTT^  T' 

wall  of  the  pipe,  and  if  the 
intensity    of     pressure     is       24" 
assumed  to  be  constant,  T  * 

is  constant  at  all  points  in  __  I  __ 

the    section.     The   sum   of 

.  -  *»• 

the  horizontal  components 

of  the  normal  pressures  acting  on  the  semicircular  segment  is  equal 
to  the  normal  pressure  on  the  vertical  projection  of  this  segment 
(Art.  29),  Calling  this  normal  pressure  P,  since  2#  =  0, 

2T=P  =  whA  =  62.  4X100X2  =  12,480  Ibs. 
and 

77=  6,240  Ibs. 


36  PRESSURE  ON   SURFACES 

The  required  thickness  of  a  steel  pipe,  using  a  safe  working 
stress  of  16,000  Ibs.  per  square  inch  is 


or  a  little  more  than       in. 


PROBLEMS 


1.  A  vertical  rectangular  gate  is  4  ft.  wide  and  6  ft.  high.     Its  upper  edge 
is  horizontal  and  on  the  water  surface.     What  is  the  total  pressure  on  the 
gate  and  where  is  the  center  of  pressure? 

2.  Solve  Problem  1  if  the  water  surface  is  5  ft.  above  the  top  of  the  gate, 
other  conditions  remaining  the  same. 

3.  Solve  Problem  2  if  the  plane  of  the  gate  makes  an  angle  of  30°  with  the 
vertical,  other  conditions  remaining  unchanged. 

4.  A  cubical  box,  24  in.  on  each  edge,  has  its  base  horizontal  and  is  half 
filled  with  water.     One  of  the  sides  is  held  in  position  by  means  of  four  screws, 
one  at  each  corner.     Find  the  tension  in  each  screw  due  to  the  water  pressure. 

6.  A  vertical,  triangular  gate  has  a  horizontal  base  4  ft.  long,  3  ft.  below 
the  vertex  and  5  ft.  below  the  water  surface.  What  is  the  total  pressure  on 
the  gate  and  where  is  the  center  of  pressure? 

6.  A  vertical,  triangular  gate  has  a  horizontal  base  3  ft.  long  and  2  ft. 
below  the  water  surface.     The  vertex  of  the  gate  is  4  ft.  below  the  base. 
What  force  normal  to  the  gate  must  be  applied  at  its  vertex  to  open  the  gate? 

7.  A  triangular  gate  having  a  horizontal  base  4  ft.  long  and  an  altitude 
of  6  ft.  is  inclined  45°  from  the  vertical  with  the  vertex  pointing  upward. 
The  base  of  the  gate  is  8  ft.  below  the  water  surface.     What  normal  force 
must  be  applied  at  the  vertex  of  the  gate  to  open  it? 

8.  A  cylindrical  tank,  having  a  vertical  axis,  is  6  ft.  in  diameter  and 
10  ft.  high.     Its  sides  are  held  in  position  by  means  of  two  steel  hoops,  one 

at  the  top  and  one  at  the  bottom.  What  is 
the  tensile  stress  in  each  hoop  when  the 
tank  is  filled  with  water? 

9.  What   is   the  greatest   height,    h,    to 
which  the  water  can  rise  without  causing  the 
dam  shown  in  Fig.  25  to  collapse?     Assume 
b  to  be  so  great  that  water  will  not  flow  over 
the  top  of  the  dam. 

10.  If  b  in  the  figure  is  20  ft.,  find  the 
least   value   of  h   at   which    the    dam    will 
collapse. 

"FIG  25  **•  ^    vertical>    triangular    gate    has    a 

horizontal  base  8  ft.  long  and  6  ft.  below 
the  water  surface.  Its  vertex  is  2  ft.  above  the  water  surface.  What  normal 
force  must  be  applied  at  the  vertex  to  open  the  gate? 

12.  A  masonry  dam  of  trapezoidal  cross-section,  with  one  face  vertical, 


PROBLEMS  37 

has  a  thickness  of  2  ft.  at  the  top  and  10  ft.  at  the  bottom.  It  is  22  ft.  high 
and  has  a  horizontal  base.  The  vertical  face  is  subjected  to  water  pressure, 
the  water  standing  15  ft.  above  the  base.  The  weight  of  the  masonry  is 
150  Ibs.  per  cubic  foot.  Where  will  the  resultant  pressure  intersect  the  base? 

13.  In  Problem  12  what  would  be  the  depth  of  water  when  the  resultant 
pressure  intersects  the  base  at  the  outer  edge  of  the  middle  third,  or  If  ft. 
from  the  middle  of  the  base? 

14.  A  vertical  triangular  surface  has  a  horizontal  base  of  4  ft.  and  an 
altitude  of  9  ft.,  the  vertex  being  below  the  base.     If  the  center  of  pressure  is 
6  in.  below  the  center  of  gravity,  how  far  is  the  base  below  the  water  surface? 

16.  Water  stands  40  ft.  above  the  top  of  a  vertical  gate  which  is  6  ft. 
square  and  weighs  3000  Ibs.  What  vertical  lift  will  be  required  to  open  the 
gate  if  the  coefficient  of  friction  between  the  gate  and  the  guides  is  0.3? 

16.  On  one  side,  water  stands  level  with  the  top  of  a  vertical,  rectangular 
gate  4  ft.  wide  and  6  ft.  high,  hinged  at  the  top;    on  the  other  side  water 
stands  3  ft.  below  the  top.     What  force  applied  at  the  bottom  of  the  gate,  at 
an  angle  of  45°  with  the  vertical,  is  required  to  open  the  gate? 

17.  A  vertical,  trapezoidal  gate  in  the  face  of  a  dam  has  a  horizontal 
base  8  ft.  below  the  water  surface.     The  gate  has  a  width  of  6  ft.  at  the  bottom 
and  3  ft.  at  the  top,  and  is  4  ft.  high.     Determine  the  total  pressure  on  the 
gate  and  the  distance  from  the  water  surface  to  the  center  of  pressure. 

18.  Determine  the  total  pressure  and  the  position  of  the  center  of  pressure 
on  a  vertical,  circular  surface  3  ft.  in  diameter,  the  center  of  which  is  4  ft. 
below  the  water  surface. 

19.  A  6-in.  pipe  line  in  which  there  is  a  90°  bend  contains  water  under  a 
gage  pressure  of  450  Ibs.  per  square  inch.      Assuming  that  the  pressure  is 
uniform  throughout  the  pipe  and  that  the  water  is  not  in  motion,  find  the  total 
longitudinal  stress  in  joints  at  either  end  of  the  bend. 


CHAPTER  IV 


IMMERSED  AND   FLOATING  BODIES 

30.  Principle  of  Archimedes. — Any  body  immersed  in  a  liquid 
is  subjected  to  a  buoyant  force  equal  to  the  weight  of  liquid  dis- 
placed. This  is  known  as  the  principle  of  Archimedes.  It  may  be 
proved  in  the  following  manner. 

The  submerged  body  ABCD  (Fig.  26)  is  referred  to  the  coor- 
dinate axes  X,  Y  and  Z.  Consider  the  small  horizontal  prism 
aia2,  parallel  to  the  X-axis,  to  have  a  cross-sectional  area  dA. 
The  X-component  of  the  normal  force  acting  on  a\  must  be  equal 

and  opposite  to  the  same  force 
acting  on  «2,  each  being  equal 
to  whdA.  There  is,  therefore, 
no  tendency  for  this  prism  to 
move  in  a  direction  parallel  to 
the  X-axis.  Since  the  same 
reasoning  may  be  applied  to 
every  other  prism  parallel  to 
a\ai  it  follows  that  there  is  no 
tendency  for  the  body  as  a 
whole  to  move  in  this  direc- 
tion. The  same  reasoning  ap- 
plies to  movement  parallel  to 
the  Z-axis  or  to  any  other  axis 
in  a  horizontal  plane.  If,  therefore,  there  is  any  tendency  for  the 
body  to  move  it  must  be  in  a  vertical  direction. 

Consider  now  the  F-components  of  the  hydrostatic  pressures 
acting  on  the  ends  of  any  vertical  prism  6162  having  a  cross- 
sectional  area,  dA,  so  small  that  the  intensity  of  pressure  on  either 
end  of  the  prism  may  be  considered  constant.  The  resultant  of 
these  pressures  will  be  the  difference  between  dP\,  the  vertical  com- 
ponent of  the  normal  pressure  at  61  equal  to  whi  dA,  acting  down- 
ward and  dP2,  the  corresponding  force  acting  at  62,  equal  to  wli?  dA, 


FIG.  26. 


CENTER   OF   BUOYANCY  39 

acting  upward.  The  resultant  pressure  will  be  upward  and  equal 
to  w(h,2  —  hi)  dA;  but  (h^  —  hi)  dA  is  the  volume  of  the  elementary 
prism  which,  multiplied  by  w,  gives  the  weight  of  the  displaced 
liquid.  Since  the  entire  body,  A  BCD,  is  made  up  of  an  infinite 
number  of  such  prisms,  it  follows  that  the  resultant  hydrostatic 
pressure  on  the  body  will  be  a  buoyant  force  equal  in  magnitude 
to  the  weight  of  the  displaced  liquid. 

If  the  weight  of  the  body  is  greater  than  the  buoyant  force 
of  the  liquid  the  body  will  sink.  On  the  other  hand,  if  the  weight 
of  the  body  is  less  than  the  buoyant  force,  the  body  will  float 
on  the  surface,  displacing  a  volume  of  liquid  having  a  weight  equal 
to  that  of  the  body. 

31.  Center  of  Buoyancy.— A  BCD  (Fig.  27)  represents  a  floating 
body.     From  the  principles  of  the  preceding  article,  the  buoyant 
force  acting  on  any  elementary  area  of  the 

submerged  surface  must  be    equal  to  the 

weight  of  the  vertical    prism  of  displaced 

liquid  directly  above  it.     Since  the  weight 

of  each  prism  is  directly  proportional  to  its 

volume,  the  center  of  gravity  of  all  these 

buoyant  forces,  or  the  center  of  buoyancy,  must  coincide  with  the 

center  of  gravity  of  the  displaced  liquid. 

32.  Stability  of  Floating  Bodies. — Any  floating  body  is  sub- 
jected to  two  systems  of  parallel  forces;   the  downward  force  of 
gravity  acting  on  each  of  the  particles  that  goes  to  make  up  the 
body  and  the  buoyant  force  of  the  liquid  acting  upward  on  the 
various  elements  of  the  submerged  surface. 

In  order  that  the  body  may  be  in  equilibrium  the  resultants 
of  these  two  systems  of  forces  must  be  colinear,  equal  and  oppo- 
site. Hence  the  center  of  buoyancy  and  the  center  of  gravity 
of  the  floating  body  must  lie  in  the  same  vertical  line. 

Fig.  28  (a)  shows  the  cross-section  of  a  ship  floating  in  an 
upright  position,  the  axis  of  symmetry  being  vertical.  For  this 
position  the  center  of  buoj^ancy  lies  on  the  axis  of  symmetry  at 
BO  which  is  the  center  of  gravity  of  the  area  ACL.  The  center  of 
gravity  of  the  ship  is  assumed  to  be  at  G.  If,  from  any  cause,  such 
as  wind  or  wave  action,  the  ship  is  made  to  heel  through  an  angle 
6,  as  shown  in  Fig.  28  (6),  the  center  of  gravity  of  the  ship  and 
cargo  remaining  unchanged,  the  center  of  buoyancy  will  shift 
to  a  new  position,  B,  which  is  the  center  of  gravity  of  the  area 


40 


IMMERSED  AND  FLOATING   BODIES 


A'C'L.  The  buoyant  force  F,  acting  upward  through  B,  and  the 
weight  of  the  ship  W,  acting  downward  through  G,  constitute  a 
couple  which  resists  further  overturning  and  tends  to  restore  the 
ship  to  its  original  upright  position.  In  all  cases,  if  the  vertical 
line  through  the  center  of  buoyancy  intersects  the  inclined  axis 
of  symmetry  at  a  point  M  above  the  center  of  gravity,  the  two 
forces  F  and  W  must  produce  a  righting  moment.  If,  however, 
M  lies  below  G  an  overturning  moment  is  produced.  The  point  M 
is  known  as  the  metacenter  and  its  distance,  GM,  from  the  center 
of  gravity  of  the  ship,  is  termed  the  metacentric  height.  The  value 
of  the  metacentric  height  is  a  measure  of  the  stability  of  the  ship. 
For  angles  of  inclination  not  greater  than  10°  or  15°  the 


FIG.  28. 

position  of  M  does  not  change  materially,  and  for  small  angles  of 
heel  the  metacentric  height  may  be  considered  constant.  For 
greater  inclinations,  however,  the  metacentric  height  varies  to  a 
greater  extent  with  the  angle  of  heel. 

33.  Determination  of  Metacentric  Height.— Fig.  29  illus- 
trates a  ship  having  a  displacement  volume,  V.  When  the  ship  is 
tilted  through  the  angle  6  the  wedge  AOA'  emerges  from  the 
water  while  the  wedge  C'OC  is  immersed.  If  the  sides  A  A'  and 
C'C  are  parallel,  these  wedges  must  be  similar  and  of  equal  volume, 
t>,  since  the  same  volume  of  water  is  displaced  by  the  ship  whether 
in  an  inclined  or  upright  position.  The  wedges  therefore  will 
have  the  same  length  and  the  water  lines  AC  and  A'C'  must  inter- 
sect on  the  axis  of  symmetry  at  0.  ^ 

When  the  ship  floats  in  an  upright  position  a  buoyant  force 


DETERMINATION  OF  METACENTRIC  HEIGHT 


41 


F,  equal  to  wv,  acts  upward  through  K,  the  center  of  gravity  of 
the  wedge  AOA'.  In  the  inclined  position  this  force  no  longer 
acts,  but  an  equal  force  F'  acts  at  K',  the  center  of  gravity  of  the 
wedge  C'OC.  It  may  be  considered  that  a  downward  force  F", 
equal  to  F,  has  been  introduced,  the  resultant  of  F"  and  F  being 
zero.  A  righting  couple  has  therefore  been  introduced  equal  to 
wvL,  L  being  the  horizontal  distance  between  the  centers  of  gravity 
of  the  wedges. 

Because  of  the  shifting  of  the  force  F  from  K  to  K'  the  line  of 


action  of  the  buoyant  force  W  acting  on  the  entire  ship  is  shifted 
from  BQ  to  B,  a  horizontal  distance  S  such  that 


=  wvL. 


(1) 


Consider  now  a  small  vertical  prism  of  the  wedge  C'OC,  at  a 
distance  x  from  0,  having  a  cross-sectional  area  dA.  The  buoyant 
force  produced  by  this  immersed  prism  is  wx  tan  6  dA,  and  the 
moment  of  this  force  about  0  is  wx2  tan  0  dA . 

The  sum  of  all  of  these  moments  for  both  wedges  must  be  equal 
to  wvL  or 

w  tan  0J  x2  dA  =  wvL  =  wVS. 

But  S  =  MB0  sin  0,  and  for  small  angles,  since  the  sine  is  very  nearly 
equal  to  the  tangent, 

Jx2  dA  =  V(MBo)  =  V(GM±GB0), 


42 


IMMERSED  AND  FLOATING   BODIES 


and  since  J  x2  dA  is  the  moment  of  inertia,  /,  of  the  water-line 
section  about  the  longitudinal  axis  through  0, 

f^TUf 1^7?  fty} 

the  sign  in  the  last  expression  being  positive  if  M  falls  below  G 

and  negative  if  above. 

Example. — Find  the  me- 
tacentric  height  of  the  rect- 
angular scow  shown  in  Fig. 
30. 

The  scow  is  40  ft.  long, 
20  ft.  wide  and  8  ft.  deep. 
It  has  a  draft  of  5  ft.  when 

FIG.  30.  floating  in  an  upright  posi- 

tion.    The  center  of  gravity 

of  the  scow  is  on  the  axis  of  symmetry,  1  ft.  above   the  water 
surface.     The  angle  of  heel  is  10°. 

The  problem  may  be  solved  by  substituting  values  in  formula 
(2),  but  the  following  method  may  be  used  conveniently  for  shapes 
such  that  the  center  of  buoyancy  can  be  readily  found. 

Since  the  center  of  buoyancy,  B,  is  at  the  center  of  gravity  of 
HA'C'F,  its  position  may  be  found  by  taking  moments  about  the 
axes  HF  and  EF.  Before  taking  moments  the  distances  KCf 
and  KF  must  be  determined. 


tan  10°  =  3.  52, 


Taking  moments  about  HF 


'  =  2.60ft. 


Taking  moments  about  EF 


£#=8.83  ft. 


PROBLEMS  43 

The  distance  of  B  from  the  inclined  axis  of  symmetry  is 
££'  =  10-8.  83  =  1.17  ft. 

From  similar  triangles 

MB'    A'K  ,.D/     1.17X20 

-BF=ra    °r    MB^-^-^-  =6.65  ft. 

The  metacentric  height  is 


=  6.  65+2.  60-6  =  3.  25  ft. 
The  righting  moment  is 

TFz  =  5X20X40X62.4X3.25sinlO* 
=  140,900  ft.-lbs. 

PROBLEMS 

1.  A  rectangular  scow  15  ft.  by  32  ft.,  having  vertical  sides,  weighs  40 
tons  (80,000  Ibs.)  .     What  is  its  draft? 

2.  If  a  rectangular  scow  18  ft.  by  40  ft.  has  a  draft  of  5  ft.  what  is  its 
weight? 

3.  A  cubic  foot  of  ice  (sp.  gr.=0.90)  floats  freely  in  a  vessel  containing 
water  whose  temperature  is  32°  F.     When  the  ice  melts,  will  the  water  level 
in  the  vessel  rise,  lower  or  remain  stationary?     Explain  why. 

4.  A  ship  of  4000  tons  displacement  floats  with  its  axis  of  symmetry 
vertical  when  a  weight  of  50  tons  is  midship.    Moving  the  weight  10  ft.  towards 
one  side  of  the  deck  causes  a  plumb  bob,  suspended  at  the  end  of  a  string  12  ft. 
long,  to  move  9  in.     Find  the  metacentric  height. 

6.  A  rectangular  scow  30  ft.  wide,  50  ft.  long,  and  12  ft.  high  has  a  draft 
of  8  ft.  Its  center  of  gravity  is  9  ft.  above  the  bottom  of  the  scow.  If  the 
scow  is  tilted  until  one  side  is  just  submerged,  determine  : 

(a)  The  position  of  the  center  of  buoyancy. 

(6)  The  metacentric  height. 

(c)  The  righting  couple,  or  the  overturning  couple. 

6.  In  Problem  5,  what  would  be  the  height  of  the  scow  (all  other  data 
remaining  unchanged)  if,  with  one  side  just  submerged,  the  scow  would  be 
in  unstable  equilibrium? 

7.  A  box,  1  ft.  square  and  6  ft.  high,  has  its  upper  end  closed  and  lower 
end  open.     By  submerging  it  vertically  with  the  open  end  down  what  is  the 
greatest  weight  the  box  can  sustain  without  sinking? 

8.  In  Problem  7,  what  weight  would  hold  the  box  in  equilibrium  with  the 
upper  end  submerged  10  ft.  below  the  surface? 

9.  A  solid  block  of  wood  (sp.  gr.  =0.6)  in  the  shape  of  a  right  cone  has  a 


44  IMMERSED  AND   FLOATING   BODIES 

base  whose  diameter  is  12  in.  and  an  altitude  of  18  in.     In  what  position 
will  this  block  float  in  water  when  it  is  in  stable  equilibrium? 

10.  A  solid  block  of  wood  (sp.  gr.=0.6)  in  the  shape  of  a  right  cylinder 
has  a  diameter  of  12  in.  and  a  length  of  15  in.  Determine  the  position  in 
which  this  block  will  float  in  water  when  in  stable  equilibrium. 


CHAPTER  V 


RELATIVE  EQUILIBRIUM  OF  LIQUIDS 

34.  Relative  Equilibrium  Defined. — In  the  preceding  chapters 
liquids  have  been  assumed  to  be  in  equilibrium  and  at  rest  with 
respect  both  to  the  earth  and  to  the  containing  vessel.     The 
present  chapter  treats  of  the  condition  where  every  particle  of  a 
liquid  is  at  rest  with  respect  to  every  other  particle  and  to  the  con- 
taining vessel,  but  the  whole  mass,  including  the  vessel,  has  a 
uniformly  accelerated  motion  with  respect  to  the  earth.     The 
liquid  is  then  in  equilibrium  and  at  rest  with  respect  to  the  vessel, 
but  it  is  neither  in  equilibrium  nor  at  rest  with  respect  to  the 
earth.    In  this  condition  a  liquid  is  said  to  be  in  relative  equilibrium. 
Since  there  is  no  motion  of  the  liquid  with  respect  to  the  vessel 
and  no  movement  between  the  water  particles  themselves  there 
can  be  no  friction. 

Hydrokinetics,  which  is  treated  in  the  following  chapters, 
deals  with  the  condition  in  which  water  particles  are  in  motion 
with  respect  to  the  earth  and  also  with  respect  to  each  other. 
In  this  case  the  retarding  effects  of  friction  must  be  considered. 

Relative  equilibrium  may  be  considered  as  an  intermediate 
state  between  hydrostatics  and  hydrokinetics. 
Two  cases  of  relative  equilibrium  will  be  dis- 
cussed. 

35.  Vessel  Moving  with  Constant  Linear 
Acceleration. — If   a  vessel  partly  filled  with 
any  liquid  moves  horizontally  along  a  straight 
line    with    a    constant    acceleration,   j,    the 
surface  of  the  liquid  will  assume  an  angle  0 
with  the  horizontal  as  shown  in  Fig.  31.     To 
determine  the  value  of  6  for  any  value   of  j, 
consider  the  forces  acting  on  a  small  mass  of 
liquid,  M,  at  any  point  0  on  the   surface. 

This  mass  is  moving  with  a  constant  horizontal  acceleration,  j, 
and  the  force   producing  the  acceleration  is  the   resultant  of  all 

45 


FIG.  31. 


46  RELATIVE  EQUILIBRIUM  OF  LIQUIDS 

the  other  forces  acting  upon  the  mass.  These  forces  are  the 
force  of  gravity,  W,  acting  vertically  downward  and  the  pres- 
sure of  all  the  contiguous  particles  of  the  liquid.  The  resultant, 
F,  of  these  forces  must  be  normal  to  the  free  surface  AB.  Since 
force  equals  mass  times  acceleration, 

P-Mj-M, (1) 

and  from  the  figure 

P  =  W  tan0 (2) 

Solving  these  two  equations  simultaneously, 

tan0  =  -?, (3) 

tf 

which  gives  the  slope  that  the  surface,  AB,  will  assume  for  any 
constant  acceleration  of  the  vessel. 

Since  0  was  assumed  to  be  anywhere  on  the  surface  and  the 
values  of  j  and  g  are  the  same  for  all  points,  it  follows  that  tan  6 
is  constant  at  all  points  on  the  surface  or,  in  other  words,  AB 
is  a  straight  line. 

The  same  value  of  8  will  hold  for  a  vessel  moving  to  the  right 
with  a  positive  acceleration  as  for  a  vessel  moving  to  the  left 
with  a  negative  acceleration  or  a  retardation. 

To  determine  the  intensity  of  pressure  at  any  point  6,  at  a 
depth  h  below  the  free  surface,  consider  the  vertical  forces  acting 
on  a  vertical  prism  ab  (Fig.  31).  Since  there  is  no  acceleration 
vertically  the  only  forces  acting  are  atmospheric  pressure  at  a, 
gravity,  and  the  upward  pressure  on  the  base  of  the  prism  at  6. 
Hence,  if  the  cross-sectional  area  is  dA, 

pi,dA=whdA -\-padA, (4) 

or 

pi,=wh+pa, (5) 

or,  neglecting  atmospheric  pressure  which  acts  throughout, 

Pb  =  wh (6) 

Therefore,  in  a  body  of  liquid  moving  with  a  horizontal  accelera- 
tion the  relative  pressure  at  any  point  is  that  due  to  the  head  of 
liquid  directly  over  the  point,  as  in  hydrostatics.  In  this  case, 
however,  it  is  evident  that  all  points  of  equal  pressure  lie  in  an 
inclined  plane  parallel  with  the  surface  of  the  liquid. 


VESSEL   ROTATING   ABOUT   A   VERTICAL  AXES  47 


In  equation  (3)  if  j  were  zero,  tan  0  would  equal  zero;  or,  in 
other  words,  if  the  vessel  were  moving  with  a  constant  velocity 
the  surface  of  the  liquid  would  be  horizontal.  Also  if  the  accelera- 
tion were  vertically  upward,  the  surface  would  obviously  be 
horizontal. 

To  determine  the  relative  pressure  at  any  point,  6,  in  a  vessel 
with  an  acceleration  upward,  consider  the 
forces  acting  on  a  vertical  prism  of  liquid  ab 
of  height  h  and  cross-sectional  area  dA  (Fig. 
32).  The  force,  P,  producing  the  acceleration 
is  the  resultant  of  all  the  forces  acting  on 
the  prism,  consisting  of  gravity  equal  to 
wh  dA,  acting  downward  and  the  static  pres- 
sure on  the  lower  end  of  the  filament  at 
6,  equal  to  pt  dA,  acting  upward.  There- 
fore 


FIG.  32. 


wh  dA. 


From  which 


=  wh+whj-.  (7) 

9 

This  shows  that  the  intensity  of 
pressure  at  any  point  within  a 
liquid  contained  in  a  vessel  hav- 
ing an  upward  acceleration,  j,  is 
greater  than  the  static  pressure 

by   an  amount   equal   to  wh~ . 

Evidently,  if  the  acceleration 
were  downward,  the  sign  of  the 
last  term  in  the  above  expression 
would  become  negative,  and  if 
j  =  g,  pb  =  0.  In  other  words,  if 
a  vessel  containing  any  liquid 
falls  freely  in  a  vacuum,  so  as 
not  to  be  retarded  by  air  friction, 
the  pressure  will  be  zero  at  all 
points  throughout  the  vessel. 

36.  Vessel  Rotating  about  a 
Vertical  Axis.— When  the  vessel  shown  in  Fig.  33  is  at  rest,  the 


FIG.  33. 


48  RELATIVE  EQUILIBRIUM   OF  LIQUIDS 

surface  of  the  liquid  is  horizontal  and  at  ran.  ra'&V  represents 
the  form  of  surface  resulting  from  rotating  the  vessel  with  a  con- 
stant angular  velocity  co  radians  per  second  about  its  vertical 
axis  OF. 

Consider  the  forces  acting  on  a  small  mass  of  liquid,  M,  at  a, 
distant  r  from  the  axis  OF. 

Since  this  mass  has  a  uniform  circular  motion  it  is  subjected  to 
a  centripetal  force, 

C  =  Ma>\    .     .     : (8) 

which  force  produces  an  acceleration  directed  toward  the  center 
of  rotation  and  is  the  resultant  of  all  the  other  forces  acting  on  the 
mass.  These  other  forces  are  the  force  of  gravity,  W  =  Mg, 
acting  vertically  downward,  and  the  pressure  exerted  by  the 
adjacent  particles  of  the  liquid.  The  resultant,'/'1,  of  this  liquid 
pressure  must  be  normal  to  the  free  surface  of  the  liquid  at  a. 

Designating  by  6  the  angle  between  the  tangent  at  a  and  the 
horizontal, 

0_dh_  C_7ifco2r_co2r 
~dr~W=  Mg  ==  g  ' 
or 

„      co2r , 
ah  =  — ar, 

\ 
which,  when  integrated  becomes,  « 


The  constant  of  integration  equals  zero,  since  when  r  equals  zero 
h  also  equals  zero. 

Since  h  and  r  are  the  only  variables  this  is  the  equation  of  a 
parabola,  and  the  liquid  surface  is  a  paraboloid  of  revolution 
about  the  F-axis.  As  the  volume  of  a  paraboloid  is  equal  to  one- 
half  that  of  the  circumscribed  cylinder  and  since  the  volume  of 
liquid  within  the  vessel  has  not  been  changed, 


The  linear  velocity  at  a  is 

v  =  ur.      ...     .....     (10) 


PROBLEMS  49 

Substituting  v  for  o>r  in  equation  (9) 


Expressed  in  words,  this  means  that  any  point  on  the  surface  of  the 
liquid  will  rise  above  the  elevation  of  greatest  depression  a  height 
equal  to  the  velocity  head  (see  Art.  43)  at  that  point. 

To  determine  the  relative  pressure  at  any  point  c  afc  a  depth  h' 
vertically  below  the  surface  at  c'  consider  the'  vertical  forces  acting 
on  the  prism  cc',  having  a  cross-sectional  area  dA.  As  this  prism 
has  no  vertical  acceleration,  2i/  =  0  and 

*  pcdA=wh'>dA 
or 

•pe  =  wh'.      .......     .'\     (12) 

* 

That  is,  the  relative  pressure  at  any  point  is  that  due  to  the 
hea*d  of  liquid  directly  over  tjie  point,  as  in  hydrostatics.  There- 
fore the  distribution  of  pressure  on  the  bottom  of  the  vessel  is 
represented  graphically  by  the  vertical  ordinates  to  the  curve 
m'b'n'r  It  also  follows  that  the  total  pressure  on  the  sides  of  the 
vessel  is  the  same  as  though  the  vessel  were  filled  to  the  level 
m'n1  and  were  not  rotating. 

4  * 

'  PROBLEMS 

1.  A  vessel  containing  water  moves  horizontally  along  a  straight  line  with 
a  constant  velocity  of  10  ft.  per  second.     What  is  the  form  of  its  water  surface? 

2.  A  vessel  partly  filled  with  water  moves  horizontally  with  a  constant, 
linear  acceleration  of  10  ft.  per  second  per  second.      What  is  the  form  of  its 
water  surface? 

3.  An  open  cylindrical  vessel,  2  ft.  in  diameter,  3  ft.  high  and  two-thirds 
filled  with  water,  rotates  about  its  vertical  axis  with  a  constant  speed  of 
QOR.PfM.     Determine: 

(a)  The  depth  of  water  at  the  center  of  the  vessel. 

(6)  The  total  pressure  on  the  cylindrical  walls. 

(c)  The  total  pressure  on  the  bottom  of  the  vessel. 

4.  In  Problem  3,  what  is  the  greatest  speed  in  revolutions  per  minute 
that  the  vessel  can  have  without  causing  any  water  to  spill  over  the  sides? 

5.  In  Problem  3,  what  speed  in  revolutions  per  minute  must  the  vessel 
have  in  order  that  the  depth  at  the  center  will  be  zero? 

6.  In  Problem  3,  what  speed  in  revolutions  per  minute  must  the  vessel 
have  in  order  that  there  may  be  no  water  within  6  in.  of  the  vertical  axis? 


50  RELATIVE   EQUILIBRIUM   OF  LIQUIDS 

7.  If  a  closed  cylindrical  vessel,  2  ft.  in  diameter,  3  ft.  high  and  completely 
filled  with  water,  rotates  about  its  vertical  axis  with  a  speed  of  240  R.P.M., 
determine  the  intensity  of  pressure  at  the  following  points: 

(a)  At  the  circumference,  just  under  the  cover. 
(6)  At  the  axis,  just  under  the  cover. 

(c)  At  the  circumference,  on  the  bottom. 

(d)  At  the  axis,  on  the  bottom. 


CHAPTER  VI 
PRINCIPLES  OF  HYDROKINETICS 

37.  Introductory. — The  principles  relating  to  the  behavior  of 
water  or  other  liquids  at  rest  are  based  upon  certain  definite  laws 
which  hold  rigidly  in  practice.  In  solving  problems  involving 
these  principles  it  is  possible  to  proceed  by  purely  rational  methods, 
the  results  obtained  being  free  from  any  doubt  or  ambiguity. 
Calculations  are  based  upon  a  few  natural  principles  which  are 
universally  true  and  simple  enough  to  permit  of  easy  application. 
In  all  problems  ordinarily  encountered  in  hydrostatics,  after  the 
unit  weight  of  the  liquid  has  been  determined,  no  other  experi- 
mental data  are  required. 

A  liquid  in  motion,  however,  presents  an  entirely  different 
condition.  Though  the  motion  undoubtedly  takes  place  in 
accordance  with  fixed  laws,  the  nature  of  these  laws  and  the 
influence  of  the  surrounding  conditions  upon  them  are  very 
complex  and  probably  incapable  of  being  expressed  in  any  exact 
mathematical  form. 

Friction  and  viscosity  affect  the  laws  of  hydrokinetics  in  a 
varying  degree  for  different  liquids.  Since  water  is  the  most 
common  liquid  with  which  the  engineer  has  to  deal  and  since,  as  a 
result,  more  is  known  about  the  laws  relating  to  the  flow  of  this 
liquid,  the  following  treatise  on  hydrokinetics  applies  only  to 
water.  The  fundamental  principles  discussed  hold  true  for  all 
liquids,  but  the  working  formulas  would  necessarily  have  to  be 
modified  for  each  different  kind  of  liquid. 

A  clearer  conception  of  the  underlying  principles  of  hydro- 
kinetics  is  made  possible  by  the  assumption  of  certain  ideal 
conditions.  This  also  permits  of  the  establishment  of  a  few 
basic  laws  which  may  be  expressed  as  fundamental  formulas. 
These  assumed  conditions,  however,  vary  widely  from  those  which 
actually  exist  and  working  formulas  based  upon  them  must  invari- 
ably be  modified  by  experimental  coefficients. 

51 


52  PRINCIPLES  OF  HYDROKINETICS 

A  formula  with  its  empirical  coefficients  included,  which 
requires  only  that  numerical  values  be  affixed  to  the  coefficients 
to  make  it  adaptable  to  the  solution  of  problems,  is  referred  to  as 
a  base  formula.  Many  formulas  used  in  hydrokinetics  differ  so 
widely  from  the  fundamental  form  that  they  have  little  if  any 
claim  to  a  rational  basis. 

During  the  last  two  centuries  many  hundreds  of  experiments 
on  flowing  water  have  been  performed.  These  experiments  have 
covered  a  wide  range  of  conditions,  and  the  data  obtained  from 
them  make  possible  the  modern  science  of  hydrokinetics. 

38.  Friction. — There  can  be  no  motion  between  two  substances 
in  contact  without  friction.     This  principle  applies  to  liquids  and 
gases  as  well  as  solids.     Water  flowing  in  any  conduit  encounters 
friction  with  the  surfaces  with  which  it  comes  in   contact.  There 
is  also  friction  between  the  moving  particles  of  water  themselves, 
commonly  called  viscosity  (Art.  6).     The  free  surface  of  water 
flowing  in  an  open  channel  encounters  the  resistance  of  the  air 
and  also  the  greater  resistance  of  the  surface  skin  which  results 
from  surface  tension  (Art.  7). 

The  amount  of  frictional  resistance  offered  by  any  surface 
increases  with  the  degree  of  roughness  of  the  surface.  The 
resistance  which  results  from  viscosity  decreases  as  the  tempera- 
ture of  the  water  increases.  The  influence  of  friction  and  viscosity 
on  the  flow  of  water  must  be  determined  experimentally. 

To  overcome  frictional  resistance  requires  an  expenditure  of 
energy.  The  expended  energy  is  transformed  into  heat.  After 
being  so  transformed  it  cannot,  through  the  ordinary  processes 
of  nature,  be  reconverted  into  any  of  the  useful  forms  of  energy 
contained  in  flowing  water  and  is  therefore  often  referred  to  as 
lost  energy. 

39.  Discharge. — The  rate  of  flow  or  the  volume  of  water  pa'ssing 
a  cross-section  of  a  stream  in  unit  time  is  called  the  discharge. 
The  symbol  Q  will  be  used  to  designate  the  discharge  in  cubic 
feet  per  second.     Other  units  of  discharge,  such  as  cubic  feet 
per  minute,  gallons  per  minute  or  gallons  per  day  are  sometimes 
employed  for  special  purposes. 

If  a  uniform  velocity  at  all  points  in  the  cross-section  of  a 
stream  were  possible  there  would  be  passing  any  cross-section  every 
second  a  prism  of  water  having  a  base  equal  to  the  cross-sectional 
area  of  the  stream  and  a  length  equal  to  the  velocity.  Because, 


STEADY  FLOW  AND   UNIFORM   FLOW  S3 

however,  of  the  varying  effects  of  friction  and  viscosity,  the 
different  filaments  of  water  move  with  different  velocities.  For 
this  reason  it  is  common  in  hydraulics  to  deal  with  mean  velocities. 
If  v  is  the  mean  velocity  in  feet  per  second  past  any  cross-section, 
and  a  is  the  cross-sectional  area  in  square  feet, 

Q  =  av       .......     .     (1) 

and 


These  simple  formulas  are  of  fundamental  importance. 

40.  Steady  Flow  and  Uniform  Flow.  —  If  the  same  quantity  of 
water  passes  any  cross-section  of  a  stream  during  equal  successive 
intervals  of  time  the  flow  is  said  to  be  steady.     If  the  quantity  of 
water  passing  any  cross-section  changes  during  successive  intervals 
of  time  the  flow  is  said  to  be  unsteady.     If  not  otherwise  stated, 
the  condition  of  steady  flow  will  be  assumed.     The  fundamental 
principles  and  formulas  based  upon  steady  flow  do  not  generally 
hold  for  unsteady  flow.    ^Problems  most  commonly  encountered 
in  practice  deal  only  with  steady  flow. 

If  in  any  reach  of  a  stream  the  velocity  at  every  cross-section 
is  the  same  at  any  instant  the  flow  is  said  to  be  uniform.  This 
condition  requires  a  stream  of  uniform  cross-section.  If  the  cross- 
section  is  not  uniform  throughout  the  reach,  in  the  portions  of  the 
reach  where  velocity  changes  occur,  the  flow  is  non-uniform. 

Thus,  uniform  flow  implies  instantaneous  similarity  of  condi- 
tions at  successive  cross-sections,  whereas  steady  flow  involves 
permanency  of  conditions  at  any  particular  cross-section. 

41.  Continuity  of  Discharge.  —  When,  at  any  instant,  the  dis- 
charge is  the  same  past  every  cross-section,  it  is  said  to  be  continu- 
ous, or  there  is  continuity  of  discharge.     The  term  continuity  of  flow 
is  also  used  to  express  this  condition.     Letting  Q,  a  and  v  represent, 
respectively,  discharge,  area  and  mean  velocity  with  similar  sub- 
scripts applying  to  the  same  cross-section,  continuity  of  discharge 
exists  when 

etc  .....     .     (3) 


Continuity  of  discharge  may  be  illustrated  by  assuming  water 
to  be  turned  into  a  canal.  At  first  there  will  be  a  greater  volume 
#f  water  flowing  near  the  entrance  than  at  sections  farther  down. 


54  PRINCIPLES  OF   HYDROKINETICS 

Under  such  conditions  the  discharge  is  not  continuous.  Ulti- 
mately, however,  if  the  supply  of  water  is  constant  and  assuming 
no  losses  from  seepage  or  evaporation,  there  will  be  the  same  quan- 
tity of  water  flowing  past  all  sections  of  the  canal,  and  the  condi- 
tion of  continuity  of  discharge  will  exist  in  the  entire  canal  regard- 
less of  whether  or  not  all  reaches  of  the  canal  have  the  same  cross- 
sectional  area.  In  a  pipe  flowing  full,  even  though  the  pipe  is 
made  up  of  several  diameters,  the  discharge  is  continuous. 

42.  Stream  Line  and  Turbulent  Motion. — Flowing  water  is 
said  to  have  stream-line  motion  if  each  particle  follows  the  same 
path  as  was  followed  by  every  preceding  particle  that  occupied  the 
same  position.  If  stream-line  motion  exists  within  a  conduit 
having  parallel  sides  the  paths  of  the  water  particles  are  parallel 
to  the  sides  of  the  conduit  and  to  each  other. 

Water  flows  with  stream-line  motion  only  at  very  low  velocities, 


(a) 

FIG.  34. 

excepting  in  very  small  pipes  where  such  motion  may  exist  at  quite 
high  velocities  (Art.  90).  Under  practically  all  conditions  encoun- 
tered in  the  field  of  engineering,  the  motion  is  turbulent,  the  water 
particles  moving  without  any  regularity  and  not  in  accordance 
with  any  known  laws.  In  Fig.  34  (a)  and  (6)  represent,  respect- 
ively, stream-line  and  turbulent  motion. 

Friction  and  viscosity  affect  the  flow  of  water  whether  the 
motion  be  stream  line  or  turbulent,  but  the  effects  produced  in  the 
two  cases  are  in  accordance  with  different  laws  (Art.  90). 

43.  Energy  and  Head. — Since  the  principles  of  energy  are 
applied  in  the  derivation  of  fundamental  hydraulic  formulas,  an 
explanation  of  such  principles  as  will  be  used  is  here  introduced. 

Energy  is  defined  as  ability  to  do  work.  Where  the  English 
system  of  units  is  employed,  both  energy  and  work  are  measured 
in  foot-pounds.  The  two  forms  of  energy  commonly  recognized 
are  kinetic  energy  and  potential  energy. 

Kinetic  energy  is  the  ability  of  a  mass  to  do  work  by  virtue  of  its 


ENERGY  AND   HEAD  55 

velocity.  Where  v  is  the  velocity  in  feet  per  second  and  M  ,  the 
mass  in  gravitational  units,  is  equal  to  W/g,  the  kinetic  energy  of 
any  mass  is  expressed  by  the  equation 


V2  V2 

which  reduces  to  —  for  a  weight  of  unity.      The  expression  —  4 
is  of  the  form 

feet  per  second  X  feet  per  second     ,. 
,    .  —  -  j  -  —  --  -5  -  =  feet, 
feet  per  second  per  second 

and  it  therefore  represents  a  linear  quantity  expressed  -in  feet. 
It  is  the  distance  which  a  body  must  fall  in  a  vacuum  to  acquire 
the  velocity  v.  When  applied  to  flowing  water  it  is  called  the 
velocity  head.  Although  representing  a  linear  quantity,  the 
velocity  head  is  directly  proportional  to  the  kinetic  energy  of  any 
mass  having  a  velocity  v  and  is  equal  to  the  kinetic  energy  of 
one  pound  of  any  substance  moving  with  a  velocity  v. 

Potential  energy  is  latent  or  potential  ability  to  do  work. 
Water  manifests  this  ability  in  two  ways: 

(a)  By  virtue  of  its  position  or  elevation  with  respect  to  some 
arbitrarily  selected  horizontal  datum  plane,  considered  in  con- 
nection with  the  action  of  gravity.  This  may  be  called  energy 
of  position,  energy  of  elevation  or  gravitational  energy. 

(6)  By  virtue  of  pressure  produced  by  the  action  of  gravity, 
or  by  the  application  of  some  external  force,  on  the  water.  This 
may  be  called  pressure  energy. 

Energy  of  position  may  be  explained  by  considering  a  mass 
having  a  weight  of  W  pounds  whose  elevation  above  any  horizontal 
datum  plane  is  h  feet.  With  respect  to  this  plane  the  mass 
has  Wh  foot-pounds  of  energy.  A  mass  weighing  one  pound 
will  have  h  foot-pounds  of  energy.  If  a  mass  weighing  one  pound 
is  placed  h  feet  below  the  datum  plane,  its  energy  with  respect 
to  the  plane  will  be  —  h  foot-pounds,  being  negative  because  this 
amount  of  energy  will  have  to  be  exerted  upon  the  mass  to  raise 
it  to  the  datum  plane  against  the  action  of  gravity.  Here  again 
the  expression  for  energy,  in  this  case  h,  represents  a  linear  quan- 
tity which  is  the  elevation  head  of  the  mass,  but  it  should  be  kept 
clearly  in  mind  that  it  is  also  the  energy  expressed  in  foot-pounds 


56  PRINCIPLES  OF  HYDROKINETICS 

contained  in  one  pound  of  water  by.  virtue  of  its  position  with 
respect  to  the  datum  plane. 

It  thus  appears  that  the  amount  of  energy  of  position  possessed 
by  a  mass  depends  upon  the  elevation  of  the  datum  plane.  In 
any  particular  problem,  however,  all  masses  should  be  referred 
to  the  same  plane.  This  gives  the  relative  amounts  of  energy 
contained  in  different  masses  or  the  relative  amounts  of  energy 
in  the  same  mass  in  different  positions,  which  is  all  that  is  usually 
required. 

The  action  of  pressure  energy  is  illustrated  by  the  piston  and 
n  cylinder  arrangement  shown 


R 


in  Fig.  35,  which  is  operated 
-p  entirely  by  water  under  a  gage 
pressure  of  p  pounds  per 
square  foot.  The  area  of  the 
piston  is  A  square  feet.  The 

cylinder  is  supplied  with  water  through  the  valve  R  and  may  be 
emptied  through  the  valve  S. 

At  the  beginning  of  the  stroke  the  piston  is  at  CD,  the  valve 
S  is  closed  and  R  is  open.  Water  enters  the  cylinder  and  slowly 
drives  the  piston  to  the  right  against  the  force  P.  Neglecting 
friction,  the  amount  of  work  done  on  the  piston  while  moving 
through  the  distance  I  feet  is  Pl  =  pAl  foot-pounds.  If  R  is  now 
closed  and  S  opened,  again  neglecting  friction,  the  piston  may  be 
moved  back  to  its  original  position  without  any  work  being  done 
upon  it.  The  quantity  of  water  required  to  do  the  work,  pAl  foot- 
pounds, is  Al  cubic  feet  and  its  weight  is  wAl  pounds.  The  amount 
of  work  done  per  pound  of  water  is  therefore 

pAl     p  f 

—r-j  =  —  foot-pounds. 

wAl    w 

Since  this  work  is  done  entirely  at  the  expense  of  pressure  energy 
and  while  the  gage  pressure  is  being  reduced  from  p  to  zero,  the 

p 
amount  of  pressure  energy  per  pound  of  water  is  ~~  foot-pounds. 

It  has  been  shown  in  Art.  17  that  —  represents  pressure  head 

or  a  linear  quantity.  If  pressure  head  is  expressed  .in  feet  of 
water  column,  it  will  also  represent  foot-pounds  of  energy  per 
pound  "of  water  as  has  been  shown  to  be  the  case  for  velocity 


BERNOULLI'S  THEOREM 


57 


head  and  elevation  head.  There  are  other  forms  of  energy,  such 
as  heat  energy  and  electrical  energy,  which  have  no  direct  bearing 
on  the  laws  governing  flowing  water. 

The  three  forms  of  energy  which  water  may  have  are  illus- 
trated in  Fig.  36.  At  A  water  is  moving  with  a  velocity  v. 
The  kinetic  energy  of  a  pound  of  water  at  A  is  v2/2g,  the  pressure 
energy  is 

£=h 

W 

and  the  energy  of  position  referred  to  the  datum  plane  MN  is  z. 
Thus,  with  respect  to  the  plane  MN  the  total  energy  per  pound  of 
water  at  any  point  A,  expressed  in  foot-pounds,  is 


(5) 


£-» 


FIG.  36. 


FIG.  37. 


44.  Bernoulli's  Theorem.— In  1738  Daniel  Bernoulli,  an 
Italian  engineer,  demonstrated  that  in  any  stream  flowing  steadily 
without  friction  the  total  energy  contained  in  a  given  mass  of 
water  is  the  same  at  every  point  in  its  path  of  flow.  In  other 
words  kinetic  energy,  pressure  energy  and  energy  of  position  may 
each  be  converted  into  either  of  the  other  two  forms,  theoretically 
without  loss.  Thus  if  there  is  a  reduction  in  the  amount  of  energy 
contained  in  any  one  form  there  must  be  an  equal  gain  in  the  sum 
of  the  other  two. 

In  Fig.  37,  bcde  represents  a  filament  of  water  flowing  with 
steady  stream-line  motion  surrounded  by  other  water  moving  with 
the  same  velocity  as  that  of  the  filament.  Under  these  conditions 


58  PRINCIPLES  OF   HYDROKINETICS 

the  frictional  loss  that  occurs  will  be  extremely  small  and  for  the 
present  it  will  be  ignored.  Every  particle  of  water  passing  the 
section  6c  will,  a  little  later,  pass  the  section  de  and  no  water  will 
pass  the  section  de  which  did  not  previously  pass  be. 

Consider  now  the  forces  acting  on  this  filament  of  water.  On 
the  section  be  whose  area  is  ai  there  is  a  normal  pressure  in  the  direc- 
tion of  flow  of  intensity  p\  producing  motion.  On  the  section  de 
whose  area  is  #2  there  is  a  normal  intensity  of  pressure  p2  parallel 
with  the  direction  of  flow  and  resisting  motion.  On  the  lateral 
surface  of  the  filament,  indicated  by  the  lines  bd  and  ce}  there 
is  a  system  of  forces  acting  normal  to  the  direction  of  motion, 
which  have  no  effect  on  the  flow  and  can  therefore  be  neglected. 
The  force  of  gravity,  equal  to  the  weight  of  the  filament,  acts 
downward.  The  work  performed  on  the  filament  by  the  three 
forces  will  now  be  investigated. 

Consider  that  in  the  time  dt  the  particles  of  water  at  be  move 
to  b'c'  with  a  velocity  v\.  In  the  same  time  interval  the  particles 
at  de  move  to  d'e'  with  a  velocity  v%.  Since  there  is  continuity  of 
flow, 


The  work,  G\,  done  by  the  force  acting  on  the  section  be  in  the 
time  dt  is  the  product  of  the  total  force  and  the  distance  through 
which  it  acts,  or 


foot-pounds  ......     (6) 

Similarly  the  work  done  on  the  section  de  is 

Gz  =  —  p2d2V2dt  foot-pounds,       ....     (7) 

being  negative  because  p2  is  opposite  in  sense  to  p\  and  resists 
motion. 

The  work  done  by  gravity  on  the  entire  mass  in  moving  from 
the  position  bcde  to  b'c'  d'e'  is  the  same  as  though  bcb'c1  were  moved 
to  the  position  ded'e'  and  the  mass  b'c'de  were  left  undisturbed. 
The  force  of  gravity  acting  on  the  mass  bcb'c'  is  equal  to  the  volume 
aividt  times  the  unit  weight  w.  If  z\  and  22  represent,  respectively, 
the  elevations  of  the  centers  of  gravity  of  bcb'c'  and  ded'e'  above 
the  datum  plane  MN,  the  distance  through  which  the  force  of 


BERNOULLI'S   THEOREM  59 

gravity  would  act  on  the  mass  bcb'c'  in  moving  it  to  the  position 
ded'e'  is  21  — 22  and  the  work  done  by  gravity  is 

G3=waiV]Ldt(zi  —  22)  foot-pounds.      .     .     .     (8) 
The  resultant  gain  in  kinetic  energy  is 

Mv22     Mv\2    waiVidt,    0 

~2 2 2i-W-»V 0) 

From  fundamental  principles  of  mechanics,  the  total  amount  of 
work  done  on  any  mass  by  any  number  of  forces  is  equal  to  the 
resultant  gain  in  kinetic  energy.  Therefore  from  equations  (6), 
(7),  (8)  and  (9). 

wa\v\dt 
2g 

Dividing  through  by  waiVidt  and  transferring,  and  remembering 
that  a\v i  =  a,2V2,  there  results 


w  w 


This  is  known  as  Bernoulli's  equation.  It  is  the  mathe- 
matical expression  of  Bernoulli's  theorem  which  is  in  reality  the 
law  of  conservation  of  energy  applied  to  flowing  water.  It  may 
be  stated  as  follows: 

Neglecting  friction,  the  total  head,  or  the  total  amount  of 
energy  per  pound  of  water,  is  the  same  at  every  point  in  the  path 
of  flow. 

Water  invariably  suffers  a  loss  of  energy  through  friction  in 
flowing  from  one  point  to  another.  If  the  direction  of  flow  is 
from  point  1  to  point  2,  the  total  energy  at  2  must  be  less  than  at  1. 
In  order  to  make  equation  (11)  balance,  a  quantity,  hf,  equal  to 
the  loss  of  energy,  or  what  is  equivalent,  the  loss  of  head  due  to 
friction  between  the  two  points,  must  be  added  to  the  right-hand 
side  of  the  equation.  Including  the  loss  of  head  due  to  friction 
Bernoulli's  equation  becomes 


This  equation  is  the  basis  of  all  rational  formulas  used  in 
hydrokinetics.     It  is  the  foundation  of  the  science. 


60 


PRINCIPLES  OF  HYDROKINETICS 


45.  Application   of  Bernoulli's   Theorem   to   Hydrostatics.— 

Although  in  hydrostatics  it  is  not  necessary  to  make  use  of  Ber- 
noulli's equation,  it  is  interesting  to  note  that  it  applies  to  water  at 
rest  as  well  as  GO  water  in  motion.  The  points  1  and  2  (Fig.  38) 
are  hi  and  fo,  respectively,  below  the  free  surface  of  a  liquid  at 
rest  and  z\  and  z2  above  the  horizontal  plane  MN.  The  liquid 
being  at  rest,  v\  and  V2  equal  zero,  and,  since  without  velocity  there 
can  be  no  friction,  hf  is  zero.  Therefore  equation  (12)  reduces  to 


or,  transposing, 
From  the  figure 


w  w 


w      w 


(13) 
(14) 


FIG.  38. 


J 
FIG.  39. 


Substituting  this  value  of  22  —  21,  equation  (14)  may  be  written 
Pi-p2  =  w(hi-h2), (15) 

which  is  the  same  as  equation  (3),  page  11. 

46.  Bernoulli's  Theorem  in  Practice. — Bernoulli's  theorem 
(Art.  44)  is  based  upon  the  assumptions  of  steady  flow,  stream-line 
motion  and  continuity  of  discharge.  Under  ordinary  conditions 
water  flows  with  turbulent  motion  (Art.  90)  whereas  stream-line 
motion  is  assumed  in  applying  Bernoulli's  theorem.  The  effect 
of  turbulence  is  to  increase  the  losses  and  therefore  this  additional 
loss  is  included  with  the  loss  of  head  due  to  friction  and  no  further 
correction  is  necessary. 

It  is  permissible  to  write  Bernoulli's  equation  between  any 
two  points  on  any  assumed  line  of  flow  provided  it  is  known  that 
there  is  flow  between  the  two  points.  Thus,  in  Fig.  39,  which 


VENTURI   METER 


61 


represents  water  discharging  from  a  reservoir  through  a  pipe, 
Bernoulli's  equation  may  be  written  between  points  A  and  B. 
The  relation  obtained  from  this  equation  is  then  assumed  to  hold 
for  each  of  the  filaments  in  the  pipe. 

47.  Venturi  Meter. — An  illustration  of  the  practical  use  of 
Bernoulli's  equation  is  provided  by  the  Venturi  meter.  This 
instrument,  which  is  used  for  measuring  the  discharge  through 
pipes,  was  invented  by  Clemens  Herschel  and  named  by  him  in 
honor  of  the  original  discoverer  of  the  principle  involved.  A 
Venturi  meter  set  in  an  inclined  position  is  illustrated  in  Fig.  40, 


FIG.  40. — Venturi  meter. 

It  consists  of  a  converging  section  of  pipe  BC  and  a  longer 
diverging  section  DE,  the  smaller  ends  being  connected  by  a 
cylindrical  section  CD,  called  the  throat.  The  larger  ends  B  and 
E,  termed  the  inlet  and  outlet,  respectively,  have  the  same  diameter 
as  the  pipe  line  in  which  the  meter  is  to  be  installed. 

Let  ai,vij  pi  and  z\  represent  the  area,  velocity,  pressure  and 
elevation,  respectively,  at  point  1  in  the  inlet.  Also  let  a2,  i»2, 
P2  and  Z2  represent  the  corresponding  quantities  at  point  2  in  the 
throat.  Writing  Bernoulli's  equation  between  points  1  and  2, 
neglecting  friction, 

vi2  ,  Pi 


62 


PRINCIPLES  OF  HYDROKINETICS 


If  piezometer  tubes  are  connected  at  the  inlet  and  throat,  the 
heights  of  water  in  these  tubes  afford  a  measure  of  the  pressure 
energy  at  the  two  points.  Since  the  relative  elevations  of  points 
1  and  2  are  also  known,  the  only  unknowns  in  equation  (16)  are 
vi  and  V2.  For  any  given  diameters  of  throat  and  inlet  the  corre- 
sponding areas  can  be  found,  and  since 


V2  can  be  expressed  in  terms  of  v\,  and  substituting  this  equivalent 
value  in  equation  (16)  leaves  v\  as  the  only  unknown.  Solving 
the  equation  for  v\  and  multiplying  the  result  by  ai  gives  the 
discharge  through  the  pipe. 

For  a  given  discharge  the  difference  between  elevations  of  water 
surfaces  in  the  two  piezometers  will  be  the  same  regardless  of 


FIG.  41. 


whether  the  meter  is  horizontal  or  inclined.  If  the  meter  is 
assumed  to  be  rotated  in  a  vertical  plane  about  point  2  until  it 
is  in  a  horizontal  position,  the  rate  of  discharge  through  the 
meter  being  unchanged,  the  total  amount  of  energy  contained 
in  the  water  at  the  inlet  must  be  the  same  as  before  the  meter  was 

rotated.     Since  v\  has  remained  constant  — +21  must  also  have 

w 

remained  constant.  The  same  reasoning  applies  to  point  2,  and 
hence  the  difference  in  elevation  of  the  water  levels  in  the  two 
piezometers  is  constant  for  all  angles  of  inclination. 

Venturi  meters  are  usually  installed  in  an  approximately 
horizontal  position. 

Example. — A  Venturi  meter  having  a  throat  4  in.  in  diameter 
is  installed  in  a  12-in.  pipe  line.  A  mercury  U-tube  connected  as 
shown  in  Fig.  41  shows  a  difference  in  height  of  mercury  columns 


*&2x 

VENTURI   METER  63 

of  9  in.,  the  remainder  of  the  tube  being  filled  with  water.  Find 
the  rate  of  discharge,  Q,  in  cubic  feet  per  second,  neglecting 
friction. 

Writing  Bernoulli's  equation  between  points  1  and  2 


Since  the  angle  of  inclination  will  not  affect  the  result,  the  meter 
can  be  assumed  to  be  horizontal.  Then  z\  and  22  cancel  and 
equation  (16)  becomes 


__==_ 
2g     2g      w      w 
From  the  given  data         ^     ^ 

E1_£?=|X13.6_  »  =9  .45. 
w     w     12  12 

The  areas  of  circles  being  proportional  to  the  squares  of  their 
diameters, 

^-I2_!-Q 

a2~42~ 
and  since  the  flow  is  continuous 

and    V2 


a* 
Substituting  these  results  in  equation  (17)  and  reducing 

5^  =  9.45 
9. 


vi  =  2 . 76  ft.  per  second. 

12X7T 

Q  =  aivi=— ~ X 2. 76  =  2. 17  cu.  ft.  per  second. 

The  pressure  in  the  throat  of  a  Venturi  meter  is  always  much 
less  than  at  the  entrance.  This  may  be  seen  from  the  following 
equation  for  a  horizontal  meter: 

vi2     pi_v22     p2 


64  PRINCIPLES  OF  HYDROKINETICS 

Since  the  velocity  increases  from  point  1  to  point  2  there  must  be 
a  corresponding  decrease  in  pressure,  otherwise  there  would  be  an 
increase  in  the  total  amount  of  energy  per  pound  of  water. 

In  the  practical  use  of  the  Venturi  meter  the  loss  of  head  due  to 
friction,  though  small,  should  not  be  neglected. 

Consider  first  the  theoretical  equation  for  the  horizontal  meter : 

v_l>Pi=v_l+P2  ( 

2g^w     2g^w 

or 

il^l=^^=h (20) 

20          w      w 

h  being  the  difference  in  pressure  at  points  1  and  2,  measured  in 
feet  of  water  column.  Since 


substituting,  (20)  becomes 

IjV-tfi2 

-  =  h (22) 


20 

This  expression  reduces  to 


(23) 


Since  equation  (23)  does  not  include  the  loss  of  energy  (or  head) 
resulting  from  friction,  it  gives  a  greater  velocity  than  is  ever 
obtained.  In  order  to  correct  the  formula  for  friction  loss  an 
empirical  coefficient,  K,  is  applied  to  it.  The  discharge  through 
the  meter  is  given  by  the  formula 


(24) 


Substituting  the  value  of  v\  given  in  equation  (23)  and  including 
the  coefficient  K,  the  formula  for  discharge  through  a  Venturi 
meter  becomes 


(25) 


PITOT  TUBE 


65 


The  value  of  K  is  affected  by  the  design  of  the  meter  and  also  by 
the  degree  of  roughness  of  its  inner  surface.  It  has  been  found 
from  experiments  that  K  usually  lies  between  0.97  and  0.99. 

48.  Pitot  Tube. — Fig.  42  illustrates  several  tubes  immersed 
vertically  in  a  stream  of  flowing  water.  The  upper  ends  of  the 
tubes  are  open  and  exposed  to  the  atmosphere.  At  the  same 
depth,  ha,  there  is  an  opening  in  each  tube  which  allows  free 
communication  between  the  tube  and  water  in  the  stream.  The 
velocity  of  the  water  at  depth  h*  is  v. 

Tubes  (a),  (b)  and  (c)  are  similar,  being  bent  through  an  angle 
of  90°,  the  tip  of  each  tube  being  open.  When  the  open  end  of 
such  a  tube  is  directed  against  the  current  as  shown  by  (a),  the 
velocity  of  the  water  causes  water  to  rise  in  the  tube  a  distance 


h  above  the  free  surface  of  the  stream.     It  is  shown  later  in  this 
article  that  h  is  equal  to  the  head  due  to  the  velocity  v. 

If  the  same  tube  is  placed  with  the  open  end  directed  down- 
stream as  in  (6)  the  pressure  at  the  opening  is  less  than  hd  and  the 
water  surface  in  the  tube  is  a  certain  distance,  hi,  below  the 
surface  of  the  stream.  A  similar  condition  exists  when  the  tube 
is  placed  with  its  lower  leg  transverse  to  the  stream  as  shown  in  (c). 
In  this  case,  according  to  experiments  by  Darcy,  the  vertical 
distance,  h%,  of  the  water  surface  in  the  tube  below  the  water 
surface  of  the  stream  is  a  little  greater  than  hi.  Similarly  for  (d), 
which  is  a  straight  tube  open  at  each  end,  there  is  a  depression, 
/i3,  of  the  water  column  in  the  tube.  The  tube  (e)  is  the  same  as 
(a)  except  that  the  tip  of  the  tube  is  closed  and  there  is  a  small 
hole  on  each  side  of  the  lower  leg.  If  this  tube  is  held  with  the 


66 


PRINCIPLES  OF   HYDROKINETICS 


lower  leg  parallel  to  the  direction  of  flow  the  water  surface  in  the 
tube  remains  at  about  the  same  elevation  as  the  water  surface  of 
the  stream,  hi,  h2  and  h%  are  all  less  than  the  velocity  head, 
but  directly  proportional  to  it.  From  experiments  by  Darcy  the 
following  approximate  values  of  hi  and  h2  were  obtained  : 


and    fc2  = 


The  conditions  of  flow  affecting  the  height  of  water  column  in 
tube  (d)  are  similar  to  those  encountered  when  piezometer  tubes 
(Art.  21)  extend  through  the  conduit  walls  into  the  stream. 
Piezometer  tubes  are  designed  to  measure  pressure  head  only  and 
in  order  that  their  readings  may  be  affected  a  minimum  amount 


he 


FIG.  43.— Pitot  tube. 


FIG.  44. 


by  the  movement  of  the  water,  their  ends  should  be  set  flush  with 
the  inner  surface  of  the  conduit  and  they  should  never  project 
beyond  this  surface. 

A  bent  L-shaped  tube  with  both  ends  open,  similar  to  Fig.  42 
(a),  is  called  a  Pilot  tube,  from  the  name  of  the  French  investigator 
who  first  used  such  a  device  for  measuring  the  velocity  of  flowing 
water. 

When  a  Pitot  tube  is  first  placed  in  the  position  shown  in 
Fig.  43,  water  enters  the  opening  at  e  until  the  water  surface 
within  the  tube  rises  a  distance  h  above  the  surface  of  the  stream. 
A  condition  of  equilibrium  is  then  established  and  the  quantity  of 
water  contained  within  the  tube  will  remain  unchanged  as  long 
as  the  flow  remains  steady.  Assuming  stream-line  motion,  the 
conditions  of  flow  near  the  entrance  to  the  tube  will  be  as  shown 
in  Fig.  44.  There  will  be  a  volume  of  dead  water  in  the  tube,  the 


PITOT  TUBE  67 

upstream  limit  of  which  is  not  definitely  known,  but  which  may 
be  represented  by  some  line  such  as  abc  or  ab'c  or  by  the  interme- 
diate line  ab"c.  Since  stream-line  motion  is  assumed,  there  must 
be  some  such  surface  of  quiescent  water  on  the  adjacent  upstream 
side  of  which  particles  of  water  will  be  moving  with  an  extremely 
low  velocity. 

Consider  a  particle  of  water  flowing  from  dtoe,d  being  on  the 
axis  of  the  tube  far  enough  upstream  so  that  the  velocity  is  not 
affected  by  the  presence  of  the  tube,  e  being  on  the  upstream  sur- 
face of  the  quiescent  water  above  referred  to  and  at  the  same 
elevation  as  d.  As  this  particle  flows  from  d  to  e  its  velocity  is 
gradually  retarded  from  Vd  to  practically  zero  at  e.  The  velocity 
head  at  e  may  therefore  be  called  zero. 

Based  on  the  above  assumptions  and  neglecting  friction, 
Bernoulli's  equation  between  the  points  d  and  e  may  be  written 


From  Fig.  43 

since  from  the  figure  he  —  hd  = 


=  he    and        -*, 

w  w 


w     w 
Substituting  in  equation  (26) 


(27) 


Since  d  is  a  point  at  any  depth,  the  general  expression  may  be 
written 


or 

v  =  V2gh  .......     (29) 

Hence  the  velocity  head  at  d  is  transformed  into  pressure  head 
at  e  and  because  of  this  increased  pressure  inside  the  tube  a  column 

v2 
of  water  will  be  maintained  whose  height  is  —  above  the  water 

level  outside. 


68 


PRINCIPLES   OF  HYDROKINETICS 


Pitot  tubes  of  the  type  shown  in  Fig.  43  are  not  practicable 
for  measuring  velocities  because  of  the  difficulty  of  determining 
the  height  of  the  water  surface  in  the  tube  above  the  surface  of  the 
stream.  In  order  to  overcome  this  difficulty  Darcy  used  an 
instrument  with  two  L-shaped  tubes  as  shown  in  Fig.  45.  One 
tube  is  directed  upstream  and  the  other  downstream,  the  two  tubes 
being  joined  at  their  upper  ends  to  a  single  tube  connected  with 
^  an  air  pump  and  provided  with  a 

stopcock  at  A.  By  opening  the 
stopcock  and  pumping  some  of 
the  air  from  the  tubes  both  water 
columns  are  raised  an  equal 
amount,  since  the  pressure  on 
their  surfaces  is  reduced  equally. 


FIG.  45. 


FIG,  46. 


The  stopcock  can  then  be  closed  and  the  difference  in  height  of 
water  columns  can  be  read. 

A  tube  of  this  kind  requires  rating,  since  the  downstream  leg, 
which  is  the  same  as  Fig.  42  (6),  does  not  measure  the  static 
pressure  of  the  water.  The  difference  in  height  of  water  columns, 

v2 
hi,  is  greater  than  the  velocity  head,  — ,  but  the  ratio  between  the 

t/ 

two  is  approximately  constant.  The  value  of  this  ratio  must  be 
determined  experimentally  for  each  instrument. 

The  tubes  shown  in  Fig.  46  give  a  difference  in  elevation  of 
water  columns  practically  equal  to  the  velocity  head.  The  leg 
measuring  static  head  is  similar  to  Fig.  42  (e). 

Pitot  tubes  may  be  rated  by  determining  their  readings  when 
placed  in  water  flowing  with  a  known  velocity  or  when  they  are 
moved  at  a  known  rate  through  still  water. 

If  the  Pitot  tube  is  used  for  measuring  velocity  in  a  closed 


PROBLEMS  69 

conduit  flowing  under  pressure,  two  tubes  are  absolutely  essential, 
one  measuring  both  pressure  and  velocity  and  the  other  measuring 
pressure  only. 

PROBLEMS 

1.  The  diameter  of  a  pipe  changes  gradually  from  6  in.  at  A  to  18  in.  at  B. 
A  is  15  ft.  lower  than  B.     If  the  pressure  at  A  is  10  Ibs.  per  square  inch  and 
at  B,  7  Ibs.  per  square  inch  when  there  are  5.0  cu.  ft.  per  second  flowing, 
determine : 

(a)  The  direction  of  flow. 

(6)  The  frictional  loss  between  the  two  points. 

2.  If  in  Problem  1  the  direction  of  flow  is  reversed,  determine  the  pressure 
at  A  if  all  other  factors,  including  the  frictional  loss,  remain  the  same. 

3.  In  Problem  1,  determine  the  diameter  of  pipe  at  B  in  order  that  the 
pressure  at  that  point  will  also  be  10  Ibs.  per  square  inch,  all  other  factors 
remaining  constant. 

4.  Determine  the  discharge  in  Problem  1,  assuming  no  frictional  loss,  all 
other  conditions  remaining  as  stated. 

5.  What  would  be  the  difference  in  pressure  in  pounds  per  square  inch 
between  A  and  B,  Problem  1,  if  there  were  6.2  cu.  ft.  per  second  flowing, 
neglecting  friction. 

6.  A  siphon  having  a  diameter  of  6  in.  throughout,  discharges  from  a 
reservoir,  A,  into  the  air  at  B.     The  summit  of  the  siphon  is  6  ft.  above  the 
water  surface  in  A  and  20  ft.  above  B.     If  there  is  a  loss  of  3  ft.  head  between 
A  and  the  summit  and  2  ft.  between  the  summit  and  B,  what  is  the  absolute 
pressure  at  the  summit  in  pounds  per  square  inch?     Also  determine  the 
rate  of  discharge  in  cubic  feet  per  second  and  in  gallons  per  day. 

7.  In  Problem  6  determine  the  absolute  pressure  in  pounds  per  square 
inch  at  the  summit  and  the  discharge  in  cubic  feet  per  second  if  the  diameter 
at  the  summit  is  5  in.  and  at  the  outlet,  6  in.,  all  other  data  remaining  the  same. 

8.  A  flaring  tube  discharges  water  from  a  reservoir  at  a  depth  of  36  ft. 
below  the  water  surface.     The  diameter  gradually  increases  from  6  in.  at  the 
throat  to  9  in.  at  the  outlet.     Neglecting  friction  determine  the  maximum 
possible  rate  of  discharge  in  cubic  feet  per  second  through  this  tube.     What  is 
the  corresponding  pressure  at  the  throat? 

9.  In  Problem  8  determine  the  maximum  possible  diameter  at  the  outlet 
at  which  the  tube  will  flow  full. 

10.  A  jet  of  water  is  directed  vertically  upward.     At  A  its  diameter  is 
3  in.  and  its  velocity  is  30  ft.  per  second.     Neglecting  air  friction,  determine 
its  diameter  at  a  point  10  ft.  above  A. 

11.  Water  is  delivered  by  a  scoop  from  a  track  tank  to  a  locomotive  tender 
that  has  a  speed  of  20  mi.  per  hour.     If  the  entrance  to  the  tender  is  7  ft. 
above  the  level  of  the  track  tank  and  3  ft.  of  head  is  lost  in  friction  at  what 
velocity  will  the  water  enter  the  tender? 

12.  In  Problem  11  what  is  the  lowest  possible  speed  of  the  train  at  which 
water  will  be  delivered  to  the  tender? 


70  PRINCIPLES  OF  HYDROKINETICS 

13.  A  Venturi  meter  having  a  diameter  of  6  in.  at  the  throat  is  installed 
in  an  18  in.  water  main.     In  a  differential  gage  partly  filled  with  mercury 
(the  remainder  of  the  tube  being  filled  with  water)  and  connected  with  the 
meter  at  the  throat  and  inlet,  the  mercury  column  stands  15  in.  higher  in  one 
leg  than  in  the  other.     Neglecting  friction,  what  is  the  discharge  through  the 
meter  in  cubic  feet  per  second? 

14.  In  Problem  13  what  is  the  discharge  if  there  is  1  ft.  of  head  lost  between 
the  inlet  and  the  throat,  all  other  conditions  remaining  the  same. 

15.  In  Problem  13  what  would  be  the  difference  in  the  level  of  the  mercury 
columns  if  the  discharge  is  5.0  cu.  ft.  per  second  and  there  is  1  ft.  of  head 
lost  between  the  inlet  and  throat? 

16.  A  Venturi  meter  is  installed  in  a  12-in.  water  main.     If  the  gage 
pressure  at  the  meter  inlet  is  8  Ibs.  per  square  inch  when  the  discharge  is  3.0 
cu.  ft.  per  second  determine  the  diameter  of  the  throat  if  the  pressure  at  that 
point  is  atmospheric.     Neglect  friction. 

17.  A  flaring  tube  discharges  water  from  a  vessel  at  a  point  10  ft.  below 
the  surface  on  which  the  gage  pressure  is  8.5  Ibs.  per  square  inch.     If  the 
diameter  of  the  throat  is  4  in.,  at  which  point  the  absolute  pressure  is  10  Ibs. 
per  square  inch,  determine  the  discharge  in  cubic  feet  per  second,  neglecting 
friction. 

18.  In  Problem  17  what  is  the  diameter  of  the  tube  at  the  discharge  end? 


CHAPTER  VII 


FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 

49.  Description  and  Definitions. — As  commonly  understood  in 
hydraulics,  an  orifice  is  an  opening  with  a  closed  perimeter  and  of 
regular  form  through  which  water  flows.  If  the  perimeter  is  not 
closed  or  if  the  opening  flows  only  partially  full  the  orifice  becomes 
a  weir  (Art.  68).  An  orifice  with  prolonged  sides,  such  as  a  piece 
of  pipe  two  or  three  diameters  in  length  set  in  the  side  of  a  reservoir, 
is  called  a  tube.  An  orifice  in  a  thick  wall  has  the  hydraulic  proper- 
ties of  a  tube.  Orifices  may  be  circular,  square,  rectangular  or  of 
any  other  regular  shape. 

The  stream  of  water  which  issues  from  an  orifice  is  termed  the 
jet.  An  orifice  with  a  sharp  upstream  edge  so  formed  that  water 
in  passing  touches  only  this  edge  is  called  a  sharp-edged  orifice. 
The  term  velocity  of  approach  as  applied  to  orifices  means  the  mean 
velocity  of  the  water  in  a  channel  leading  up  to  an  orifice,  The 
portion  of  the  channel  where  the  velocity 
of  approach  is  considered  to  occur  is 
designated  the  channel  of  approach.  An 
orifice  is  spoken  of  as  a  vertical  or  hori- 
zontal orifice  depending  upon  whether  it 
lies  in  a  vertical  or  horizontal  plane. 

60.  Characteristics  of  the  Jet. — Fig. 
47  represents  a  sharp-edged,  circular 
orifice.  The  water  particles  approach 
the  orifice  in  converging  directions  as 
shown  by  the  paths  in  the  figure. 
Because  of  the  inertia  of  those  particles 
whose  velocities  have  components  par- 
allel to  the  plane  of  the  orifice,  it  is 
not  possible  to  make  abrupt  changes  in 
their  directions  the  instant  they  leave 
the  orifice  and  they  therefore  continue  to  move  in  curvilinear 

71 


FIG.  47.— Sharp-edged 
orifice. 


72       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 

paths,  thus  causing  the  jet  to  contract  for  some  distance  beyond 
the  orifice.  This  phenomenon  of  contraction  is  referred  to  as  the 
contraction  of  the  jet  and  the  section  where  contraction  ceases  is 
called  the  vena  contracta.  The  vena  contracta  has  been  found  to  be 
at  a  distance  equal  to  about  one-half  the  diameter  of  the  orifice 
from  the  plane  of  the  orifice,  at  a  in  figure. 

Beyond  the  vena  contracta  the  cross-sectional  area  of  the  jet 
does  not  undergo  any  change  excepting  insofar  as  it  is  affected  by 
gravity.  If  the  direction  of  the  jet  is  vertically  upward  (Fig.  51), 
or  if  it  has  an  upward  component,  the  force  of  gravity  retards  its 
velocity  and  thus  increases  its  cross-sectional  area;  and,  con- 
versely, if  »the  direction  of  the  jet  has  a  downward  component, 
gravity  increases  its  velocity  and  decreases  its  cross-sectional 
area. 

If  a  jet  discharges  into  the  air  the  pressure  within  the  jet  at 
its  vena  contracta  and  beyond  is  atmospheric  pressure.  This 
may  be  seen  by  investigating  the  conditions  which  will  ^esult 
from  assuming  pressures  greater  or  less  than  atmospheric  pressure. 
If,  for  example,  the  pressure  within  a  cross-section  is  assumed  to 
be  greater  than  atmospheric  pressure,  there  will  be  an  unbalanced 
pressure  along  every  radius  of  the  section — that  is,  the  pressure 
at  the  center  will  be  greater  than  at  the  circumference.  Since 
water  is  incapable  of  resisting  tensile  stress  this  would  cause  the 
jet  to  expand.  In  a  similar  manner  if  the  internal  pressure  is 
assumed  to  be  less  than  atmospheric,  since  water  unconfined  is 
incapable  of  resisting  a  compressive  force,  the  unbalanced  pressure 
will  produce  an  acceleration  and  therefore  a  further  contraction. 
Since  neither  expansion  nor  contraction  occurs,  it  follows  that  the 
pressure  throughout  the  vena  contracta  must  be  atmospheric 
pressure. 

Between  the  plane  of  the  orifice  and  the  vena  contracta  the 
pressure  within  the  jet  is  greater  than  atmospheric  pressure 
because  of  the  centripetal  force  necessary  to  change  the  direction 
of  motion  of  the  particles.  That  this  pressure  must  be  greater 
than  atmospheric  can  easily  be  proved  by  writing  Bernoulli's 
equation  between  a  water  particle  in  the  jet  back  of  the  vena 
contracta  and  another  particle  in  the  vena  contracta. 

The  form  assumed  by  jets  issuing  from  orifices  of  different 
shapes  presents  an  interesting  phenomenon.  The  cross-section 
of  the  jet  is  similar  to  the  shape  of  the  orifice  until  the  vena  con- 


FUNDAMENTAL    ORIFICE    FORMULA 


73 


o 


tracta  is  reached.  Fig.  48  shows  various  cross-sections  of  jets 
issuing  from  square,  triangular  and  elliptical  orifices.  The  left- 
hand  diagram  in  each  case  is  a 
cross-section  of  the  jet  near  the 
vena  contracta.  The  following 
diagrams  are  cross-sections  at  suc- 
cessively greater  distances  from  the 
orifice.  This  phenomenon,  which 
is  common  to  all  shapes  of  orifices 
excepting  circular  orifices,  is  known 
as  the  inversion  of  the  jet.  After 
passing  through  the  fourth  stage 


A    O 


Y 


0     O 

shown  in  the  figure  the  jet  reverts  ^  ^         ,  .  .  . 

&  FIG.  48. — Form  of  jet  from  square, 

to  its  original  form  and  continues      triangular  and  elliptical  orifices, 
to  pass  through  the  cycle  of  changes 

described  above  as  long  as  it  flows  freely  or  is  not  broken  up  by 
wind  or  air  friction. 

51.  Fundamental  Orifice  Formula. — Fig.  49    represents    the 
general  case  of  water  discharging  through  an  orifice.     In  the 


,   FIG.  49. — Discharge  from  orifice. 

derivation  of  the  fundamental  formula  it  will  be  assumed  that  the 
water  flows  without  friction  and  also  that  there  is  no  contraction 
of  the  jet  and  therefore  no  pressure  within  the  jet  in  the  plane 
of  the  orifice.  In  order  to  write  a  general  expression  applicable 
to  all  filaments,  it  will  be  necessary  to  make  the  further  assump- 
tion that  all  of  the  water  particles  in  a  cross-section  of  the 
channel  of  approach  flow  with  the  same  velocity. 

There  are  two  chambers,  A  and  B,  the  gas  pressure  in  chamber 
A  being  pA  and  in  chamber  B  being  pB,  the  relation  of  pA  to  pB 


74       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 

being  such  that  water  will  flow  from  chamber  A  to  chamber  B. 
The  flowing  water  may  be  considered  to  be  made  up  of  filaments 
of  which  mn  is  one,  m  being  a  point  in  the  water  in  chamber  A 
and  n  a  point  in  the  jet  in  the  plane  of  the  orifice.  The  filament 
passes  through  the  orifice  at  a  distance  h  below  the  surface  of  the 
water.  The  point  m  is  at  a  distance  hm  below  the  water  surface 
and  at  a  distance  z  above  n.  vm  is  the  velocity  at  m  and  vn  is 
the  velocity  at  n.  Bernoulli's  equation  may  be  written  between 
the  points  m  and  n  as  follows : 


and  since  hm-}-z  = 


and 


These  formulas  are  general  expressions  of  relation  between  velocity 
and  head  for  any  filament. 

Since  the  filaments  at  different  elevations  discharge  through 
a  vertical  orifice  under  different  heads  their  velocities  are  not 
the  same.  Where,  however,  the  head  is  large  in  comparison  with 
the  height  of  the  opening,  the  mean  velocity  of  the  jet  may  be 
taken  as  the  velocity  due  to  the  mean  head.  The  theoretical  mean 
velocity  thus  obtained  may  be  represented  by  the  symbol  vt. 
Introducing  also  the  assumption  that  all  of  the  water  particles 
in  a  cross-section  of  the  channel  of  approach  flow  with  the  same 
velocity,  V;  vn  and  vm  in  formulas  (2)  and  (3)  may  be  replaced, 
respectively,  by  vt  and  V,  which  gives 


2g     2g     w      w 
and 


From  the  definition  (Art.  49)  V  is  the  velocity  of  approach. 
The  condition  most  commonly  encountered  is  that  illustrated 


FUNDAMENTAL    ORIFICE    FORMULA 


73 


in  Fig.  50,  where  the  surface  of  the  water  and  the  jet  are  each 
exposed  to  the  atmosphere.  In  this  case  pA  =  pB  =  atmospheric 
pressure  and  formulas  (4)  and  (5)  reduce,  respectively,  to 


(6) 


and 


If  the  cross-sectional  area 
of  the  reservoir  or  chan- 
nel leading  up  to  the 
orifice  is  large  in  com- 
parison with  the  area  of 
the  orifice  the  velocity  of 
approach,  V,  may  be 
called  zero  and  equations 
(6)  and  (7)  reduce,  respect- 
ively, to 


and 


FIG.  50. — Orifice  with  water  surface  and 

jet  subjected  to  equal  pressures. 
,,2 


(8) 


(9) 

These  formulas  express  the  theoretical  relation  between  head  and 
velocity  for  an  orifice  discharging  from  a  relatively  large  body  of 

water  whose  surface  is  subjected 
to  the  same  pressure  as  the  jet. 
It  is  under  this  condition  that 
discharge  from  orifices  ordinarily 
occurs  and  the  above  formulas 
are  the  ones  most  commonly 
used.  Since  these  formulas  also 
express  the  relation  between 
potential  head  and  velocity  head 
(Art.  43) -they  have  a  wide  appli- 
cation in  hydraulics. 

Formula  (9)  is  also  the  formula 
for  the  velocity  acquired  by  a 
body  falling  a  distance  h  through  space.  The  theoretical  velocity 
of  water  flowing  through  an  orifice  is  therefore  the  velocity  acquired 


FIG.  51. — Horizontal  orifice. 


76       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 

by  a  body  falling  freely  in  vacuo  through  a  distance  equal  to  the 
head  on  the  orifice.  This  principle,  discovered  by  Torricelli  in 
1644,  is  known  as  Torricelli's  theorem.  Fig.  51  illustrates  a  hori- 
zontal orifice  discharging  under  a  head  h.  According  to  Torri- 
celli's theorem  the  jet  should  rise  to  a  height  h,  but  experiments 
show  that  the  actual  height  to  which  the  jet  rises  is  slightly  less 
than  h.  The  discrepancy  is  due  to  the  retarding  effects  of  friction 
and  viscosity.  This  matter  is  discussed  more  fully  in  the  following 
article. 

52.  Orifice  Coefficients. — The  assumptions  which  were  made 
in  the  derivation  of  formula  (5)  may  be  summarized  briefly  as 
follows : 

(a)  All  water  particles  in  a  cross-section  of  the  channel  of 
approach  flow  with  the  same  velocity. 
(6)  There  is  no  contraction  of  the  jet. 
(c)  The  water  flows  without  friction. 

Since  these  conditions  do  not  in  reality  exist,  it  is  necessary 
to  modify  the  derived  formulas  to  make  them  applicable  to  actual 
conditions.  To  accomplish  this,  three  empirical  coefficients  are 
applied  to  formula  (5),  there  being  one  coefficient  to  correct  for 
the  difference  between  the  assumed  conditions  and  the  actual  con- 
ditions for  each  of  the  above  assumptions.  The  method  of  cor- 
recting for  each  assumption  will  be  discussed  in  the  order  given 
above. 

(a)  Correction  for  non-uniformity  of  velocity  in  cross-section  of 
channel  of  approach.  The  effect  of  the  variation  in  velocity  in  a 
cross-section  of  the  channel  of  approach — that  is,  the  variation 
in  the  velocity  with  which  the  water  particles  in  the  different 
filaments  approach  the  orifice,  is  similar  to  the  effect  of  this  condi- 
tion on  the  discharge  over  weirs.  The  matter  being  of  relatively 
much  greater  importance  in  this  connection  is  taken  up  under 
weirs  and  will  not  be  discussed  here.  (See  Art.  72  (a) .) 

The  commonly  accepted  method  of  modifying  formula  (5)  so 
as  to  have  it  include  this  correction  is  to  apply  a  coefficient  a  to 

V2 
the  term  — .     The  value  of  a  has  not  been  determined  for  orifices. 

It  varies  with  the  distribution  of  velocities  in  the  channel  of  ap- 
proach and  is  always  greater  than  unity.  With  the  coefficient 


ORIFICE   COEFFICIENTS  77 

a  included,  calling  vf  the  velocity  after  the  correction  has  been 
applied,  formula  (5)  becomes 


(b)  Correction  for  contraction. — The  ratio  of  the  cross-sectional 
area  of  the  jet  at  the  vena  contracta  to  the  area  of  the  orifice  is 
called  the  coefficient  of  contraction.  Thus,  if  a'  and  a  are,  respect- 
ively, the  cross-sectional  area  of  the  jet  at  the  vena  contracta 
and  the  area  of  the  orifice  and  Cc  is  the  coefficient  of  contraction, 

.n      o!  , 

Cc  =  —     or     a'  =  Cca. 

If  v  is  the  actual  mean  velocity  in  the  vena  contracta  the  discharge 
through  the  orifice  is 

en) 


The  mean  value  of  Cc  is  about  0.62.      It  varies  slightly  with  the 
head  and  size  of  orifice. 

(c)  Correction  for  friction. — The  velocity  of  the  jet  suffers  a 
retardation  due  to  the  combined  effects  of  friction  and  viscosity. 
The  ratio  of  the  actual  mean  velocity,  v,  to  the  velocity,  v',  which 
would  exist  without  friction,  has  been  termed  the  coefficient  of 
velocity,  but  it  might  more  properly  be  called  the  coefficient  of 
friction.  Designating  the  coefficient  of  velocity  by  the  sumbol  Cv 

Cv  =  —    or    v  =  Cvv'. 

The  average  value  of  Cv  for  a  sharp-edged  orifice  is  about  0.98. 

Substituting  the  value  of  v'  given  in  formula  (10),  the  general 
formula  for  mean  velocity  of  a  jet  issuing  from  an  orifice,  with  the 
two  coefficients  to  correct,  respectively,  for  friction  and  the 
assumption  of  uniform  velocity  in  a  cross-section  of  the  channel 
of  approach,  becomes 


If  the  pressures  pA  and  pB  are  equal 

.    . (13) 


78       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 

If  the  velocity  of  approach  is  so  small  that  it  can  be  called  zero 
without  introducing  an  appreciable  error 

v  =  C0V2^h.        .     .• (14) 

Substituting  the  value  of  v  given  by  formula  (12)  in  (11) 
the  general  formula  for  discharge  through  an  orifice  with  the  three 
corrective  coefficients  becomes 


.     .     .(15) 

It  is  usual  to  combine  CCCV  into  a  single  coefficient,  C,  called 
the  coefficient  of  discharge.  Substituting  C  for  CCCV,  the  general 
formula  for  discharge  is 


....     (16) 

u/         / 

If  the  pressures  pA  and  pB  are  equal 

\ 

(17) 


If  the  velocity  of  approach  V  is  so  small  that  it  can  be  considered 
to  equal  zero 


(18) 


As  orifices  are  ordinarily  used,  the  pressures  pA  and  pB  are  equal 
and  the  velocity  of  approach  is  so  small  that  it  may  be  neglected 
without  appreciable  error.  Formula  (18)  is  therefore  recognized 
as  the  common  discharge  formula  for  an  orifice. 

In  the  remainder  of  this  chapter,  if  not  otherwise  specified,  it 
will  be  assumed  that  the  pressure  on  the  water  surface  is  the  same 
as  the  pressure  on  the  jet,  and  the  velocity  of  approach  will  be 
considered  to  be  so  small  as  to  be  negligible.  Formulas  (14) 
and  (18)  then  become  the  respective  formulas  for  mean  velocity 
and  discharge. 

Numerical  values  of  Cc  and  Cc  may  be  obtained  experimentally 
but  an  accurate  determination  is  extremely  difficult.  Cc  may  be 
obtained  approximately  by  measuring  the  diameters  of  the  vena 
contracta  and  orifice  with  calipers,  the  coefficient  of  contraction 
being  equal  to  the  ratio  of  the  squares  of  their  respective 


ORIFICE   COEFFICIENTS 


79 


COEFFICIENTS  OF  DISCHARGE  (C)  FOR  CIRCULAR  ORIFICES 


Head 

h 

in  feet 

Diameter  of  orifice  in  feet 

0.02 

0.04 

0.07 

0.1 

0.2 

0.6 

1.0 

0.4 

0.637 

0.624 

0.618 

0.6 

0.655 

.630 

.618 

.613 

0.601 

0.593 

0.8 

.648 

.626 

.615 

.610 

.601 

.594 

0.590 

1.0 

.644 

.623 

.612 

.608 

.600 

.595 

.591 

1.5 

.637 

.618 

.608 

.605 

.600 

.596 

.593 

2.0 

.632 

.614 

.607 

.604 

.599 

.597 

.595 

2.5 

.629 

.612 

.605 

.603 

.599 

.598 

.596 

3.0 

.627 

.611 

.604 

.603 

.599 

.598 

.597 

4.0 

.623 

.609 

.603 

.602 

.599 

.597 

.596 

6.0 

.618 

.607 

.602 

.600 

.598 

.597 

.596 

8.0 

.614 

.605 

.601 

.600 

.598 

.596 

.596 

10.0 

.611 

.603 

.599 

.598 

.597 

.596 

.595 

20.0 

.601 

.599 

.597 

.596 

.596 

.596 

.594 

50.0 

.596 

.595 

.594 

.594 

.594 

.594 

.593 

100.0 

.593 

.592 

.592 

.592 

.592 

.592 

.592 

COEFFICIENTS  OF  DISCHARGE     (C)  FOR  SQUARE  ORIFICES 


Head 
h 
in  feet 

Side  of  square  in  feet 

0.02 

0.04 

0.07 

0.1 

0.2 

0.6 

1.0 

0.4 

0.643 

0.628 

0.621 

0.6 

0.660 

.636 

.623 

.617 

0.605 

0.598 

0.8 

.652 

.631 

.620 

.615 

.605 

.600 

0.597 

1.0 

.648 

.628 

.618 

.613 

.605 

.601 

.599 

1.5 

.641 

.622 

.614 

.610 

.605 

.602 

.601 

2.0 

.637 

.619 

.612 

.608 

.605 

.604 

.602 

2.5 

.634 

.617 

.610 

.607 

.605 

.604 

.602 

3.0 

.632 

.616 

.609 

.607 

.605 

.604 

.603 

4.0 

.628 

.614 

.608 

.606 

.605 

.603 

.602 

6.0 

.623 

.612 

.607 

.605 

.604 

.603 

.602 

8.0 

.619 

.610 

.606 

.605 

.604 

.603 

.602 

10.0 

.616 

.608 

.605 

.604 

.603 

.602 

.601 

20.0 

.606 

.604 

.602 

.602 

.602 

.601 

.600 

50.0 

.602 

.601 

.601 

.600 

.600 

.599 

.599 

100.0 

.599 

.598 

.598 

.598 

.598 

.598 

.598 

80       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 

diameters.  A  Pitot  tube  may  be  used  to  determine  approximately 
velocities  in  the  vena  contracta. 

The  coefficient  of  discharge  may  be  obtained  with  great 
accuracy  by  measuring  the  quantity  of  water  flowing  from  an 
orifice  of  known  dimensions  in  a  given  time  and  determining  the 
ratio  between  this  discharge  and  the  theoretical  discharge.  Since 
in  practice  it  is  usually  the  discharge  from  orifices  that  is  required, 
it  is  the  coefficient  of  discharge  that  is  of  greatest  value  to  en- 
gineers. An  average  value  of  the  coefficient  of  discharge  is  about 
0.60.  It  is  not  a  constant,  but  varies  with  the  head  and  also  with 
the  shape  and  size  of  the  opening.  On  page  79  are  tables  of 
values  of  C  for  circular  and  square  orifices  taken  from  Hamilton 
Smith's  Hydraulics.  Sharp-edged  orifices  provide  an  accurate 
means  of  measuring  small  rates  of  discharge. 

53.  Algebraic  Transformation  of  Formula  with  Velocity  of 
Approach  Correction. — The  fundamental  orifice  formula  with 
velocity  of  approach  correction  as  derived  in  Art.  52  is 


(17) 


By  definition  V  =  Q/A  where  A  is  the  cross-sectional  area  of  the 
stream  in  the  channel  of  approach.  Substituting  this  value  of  V 
and  reducing,  equation  (17)  becomes 


CaV2gh 

^  I — o* 


(19) 


Expanding  the  denominator  by  the  binomial  theorem  gives  a 
diminishing  series  and  dropping  all  terms  excepting  the  first 
two  since  they  will  be  very  small  quantities,  the  formula  may  be 
expressed  in  the  approximately  equivalent  form 

.     (20) 


The  term  within  the  parenthesis  is  the  velocity  of  approach 
corrective  factor.  It  becomes  unity  when  the  ratio  of  the  orifice 
to  the  cross-sectional  area  of  the  stream  in  the  channel  of  approach 
is  so  small  that  it  may  be  considered  zero.  Where  a  correction 
for  velocity  of  approach  is  required,  formula  (20)  will  be  found 
more  convenient  than  formula  (17). 


HEAD   LOST  IN  AN  ORIFICE 


81 


54.  Head  Lost  in  an  Orifice. — Consider  water  to  be  discharging 
from  an  orifice  under  a  head  h  (Fig.  52).  Because  of  friction,  the 
velocity  of  discharge  will  be  less  than  V2gh  or,  from  formula  (14), 
page  78 

......         (14) 


The  head  producing  discharge 
is  therefore, 

/»=  —  - 

C2  2g' 

That   is,  h  is  the   total  head, 
including  the  lost  head.     The 


(19) 


FIG.  52. — Sharp-edged  orifice. 


FIG,  53.— Path  of  jet. 


head   that  is  not  lost  is  the  velocity  head   due  to  the   actual 
velocity  v.     Therefore, 

Lost  head  =  Total  head  —  velocity  head  or,  placing  the  symbol 
ho  for  lost  head, 

v2 

•     '     '     •     (22) 


(23) 


Formulas  (22)  and  (23)  are  fundamental  and  are  applicable 
to  any  orifice  or  tube  whose  coefficient  of  velocity  is  known. 


v2 
For  a  sharp-edged  orifice,  since  Cc  =  0.98,  h0  =  0.041  — . 

Since  v  =  Cvvt,  formula  (22)  reduces  to 


and  substituting  the  value  of  Ct  for  a  sharp-edged  orifice 

h. 


82       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 


55.  Path  of  Jet. — When  water  issues  from  an  orifice  the 
direction  of  the  jet  is  at  first  normal  to  the  plane  of  the  orifice 
but,  for  orifices  not  in  a  horizontal  plane,  the  force  of  gravity 
causes  it  immediately  to  begin  to  curve  downward.  Let  x  (Fig. 
53)  be  the  abscissa  and  y  the  ordinate  of  any  point  in  the  path 
of  a  jet  discharging  from  a  vertical  orifice.  The  space  x  will  be 
described  uniformly  in  a  certain  time  t  and  if  v  is  the  velocity 
with  which  water  leaves  the  orifice 

x  =  vt. 

The  jet  has  a  downward  acceleration  which  conforms  to  the  law  of 
falling  bodies  and  therefore 

^ 
y~  2' 

Eliminating  t  between  the  two  equations 


(24) 


which  is  the  equation  of  a  parabola  with  its  vertex  at  the  orifice. 
Since  by  formula  (14),  page  78, 

v  =  CvV~2gh, (14) 

equation  (24)  may  also  be  written 

x2  =  4C*hy.  .  (25) 


This  formula  indicates  an  experimental  method  of  obtaining  C»; 

x,  y  and  h  may  be  measured  and  substituted  in  the  formula  and 

Cv  may  be  computed 

56.  Orifices  under  Low  Heads. — Where  the  head  on  a  vertical 

orifice  is  small  in  comparison  with  the  height  of  the  orifice  there  is 

theoretically  an  appreciable  dif- 
ference between  the  discharge 
obtained  by  assuming  the  mean 
velocity  to  be  that  due  to  the 
mean  head  and  the  discharge  ob- 
tained by  taking  into  considera- 
tion the  variation  in  head.  The 

FIG.  54.-Rectangular  orifice.          6XaCt    formula    f°r    octangular 

orifices  is  derived  as  follows: 
Fig.  54   shows  &  rectangular  orifice  of  width  L  and  height  D, 


ORIFICES   UNDER  LOW  HEADS  83 

the  water  surface  and  jet  being  each  subjected  to  atmospheric 
pressure,  hi  and  hz  are  the  respective  heads  on  the  upper  and 
lower  edges  of  the  orifice.  Neglecting  velocity  of  approach,  the 
theoretical  discharge  through  any  elementary  strip  of  area  Ldy 
at  a  distance  y  below  the  water  surface,  is  given  by  the  equation 


which  integrated  between  the  limits  of  h%  and  hi  gives 

-hiK)  ......     (26) 


This  formula  gives  the  theoretical  discharge  from  an  orifice 
where  the  pressures  on  the  water  surface  and  jet  are  equal  and 
the  velocity  of  approach  is  considered  to  be  zero.  When  hi  is 
zero  —  that  is,  when  the  water  surface  does  not  touch  the  upper 
edge  of  the  opening,  the  formula  reduces  to 

,       ......     (27) 


which  is  the  theoretical  formula  for  discharge  over  a  weir  without 
velocity  of  approach  correction  (see  Art.  70). 

To  make  formula  (26)  applicable  to  actual  conditions  a  coeffi- 
cient of  discharge  must  be  introduced  and  the  formula  becomes 

-hiK)  ......     (28) 


Values  of  C  for  this  formula  have  not  been  well  determined  and 
it  is  seldom  used  in  practice.  Formula  (18),  which  for  rectangular 
orifices  may  be  written 

.......     (29) 


h  being  the  head  on  the  center  of  the  orifice  and  C  the  coefficient 
of  discharge  for  rectangular  orifices  as  given  on  page  79,  may 
be  used  satisfactorily  even  for  quite  low  heads  since  these  values 
of  C  include  corrections  for  the  approximations  contained  in  the 
formula. 

The  theoretical  difference  between  formulas  (28)  and  (29) 
may  be  shown  as  follows  :  h  being  the  head  on  the  center  of  the 
rectangle,  h2  =  h-\-^D  and  hi  =  h  —  %D.  Substituting  these  values 
in  equation  (26)  and  expanding  them  by  the  binomial  theorem, 

.      (30) 


84       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 


This  shows  that  formula  (29)  always  gives  a  greater  discharge  than 
formula  (28)  if  the  same  value  of  C  is  used  in  each  case.  For 
h  =  D,  the  sum  of  the  infinite  series  is  0.989  and  for  h  =  2D,  it 
is  0.997.  The  theoretical  error  introduced  by  using  formula  (29) 
is  thus  about  1  percent  where  h  =  D  and  0.3  of  1  per  cent  where 
h  =  2D. 

In  a  manner  similar  to  that  described  above  for  rectangular 
orifices,  the  discharge  for  a  circular  orifice  may  be  shown  to  be 


FIG.  55. — Orifice  with  bottom  con- 
traction partially  suppressed. 


FIG.  56. — Orifice  with  bottom  con- 
traction completely  suppressed. 


in  which  D  is  the  diameter  of  orifice  and  h  is  the  head  on  the  center 
of  orifice.  This  formula  gives  results  differing  from  those  obtained 
by  the  approximate  formula  similar  to  the  corresponding  formulas 
for  rectangular  orifices.  If  h  =  D  the  sum  of  the  series  is  0.992 
and  if  h  =  2D  the  sum  is  0.998.  Formula  (31)  is  seldom,  if  ever, 
used  in  practice. 

57.  Suppression  of  Contraction. — The  effect  of  constructing  an 
orifice  so  as  to  reduce  the  contraction  is  to  increase  the  cross- 
sectional  area  of  the  jet  and  thus  to  increase  the  discharge.  If  an 
orifice  is  placed  close  to  a  side  or  the  bottom  of  a  reservoir  the 
tendency  of  the  filaments  of  water  to  approach  the  orifice  from  all 
directions  (Fig.  55)  is  restricted  and  some  of  the  filaments  must 
approach  in  a  direction  more  nearly  parallel  to  the  direction  of 
the  jet  than  they  would  otherwise.  If  the  orifice  is  flush  with 
one  side  or  the  bottom  (Fig.  56)  the  contraction  on  that  side 
of  the  orifice  will  be  wholly  suppressed. 


STANDARD   SHORT  TUBE 


85 


In  a  similar  manner,  rounding  the  inner  edge  of  the  orifice 
(Fig.  57)  reduces  contraction.  An  orifice  constructed  to  con- 
form to  the  shape  of  the  jet  which  issues  from  a  sharp-edged  orifice 


FIG.  57. — Orifice  with  rounded  entrance.        FIG.  58. — Bell-mouth  orifice. 


FIG.  59. — Sharp-edged  orifice 
with  extended  sides. 


FIG.  60. — Standard  short  tube. 


(Fig.  58)  is  called  a  bell-mouth  orifice.      The  coefficient   of  con- 
traction of  such  an  orifice  approaches  very  close  to  unity. 

58.  Standard  Short  Tube. — Extending  the  sides  of  an  orifice 
does  not  affect  the  discharge  so  long  as  the  jet  springs  clear.     The 


86       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 


orifice  illustrated  in  Fig.  59  has  a  sharp  upstream  corner  and  the 
conditions  of  flow  are  the  same  as  for  a  sharp-edged  orifice  in  a 
thin  plate. 

When  the  jet  touches  the  sides  of  the  orifice  the  conditions  of 
flow  are  changed.  A  circular  orifice  with  a  sharp  edge  having 
sides  extended  to  about  1\  diameters  is  called  a  standard  short  tube. 
The  jet  is  contracted  by  the  edge  of  the  orifice,  as  at  m  (Fig.  60), 
and  for  low  heads  it  will  expand  and  fill  the  tube.  For  high  heads 
the  jet  may  at  first  spring  clear  of  the  sides  of  the  tube,  but  by 
temporarily  stopping  the  tube  at  its  discharging  end  and  allowing 
the  water  to  escape,  the  tube  can  be  made  to  flow  full.  The  moving 
water  carries  with  it  a  portion  of  the  air  which  is  entrapped  in  the 
space,  s,  causing  a  pressure  less  than  atmospheric  pressure.  The 
result  is  to  increase  the  head  under  which  water  enters  the  orifice 
and  therefore  the  discharge  is  greater  than  occurs  from  a  sharp- 
edged  orifice  of  the  same  diameter  discharging  freely  into  the  air. 

Conditions  at  the  outlet 
end  of  the  tube  will  first  be 
considered.  By  writing  Ber- 
noulli's equation  between  a 
point  in  the  reservoir  where 
the  velocity  of  approach  may 
be  considered  zero  and  a 
point,  n,  in  the  outlet  (Fig.  61) 
there  is  obtained,  as  for  an 
orifice  (Art.  51)  the  relation 


V,' 


(9) 


Since  the  tube  flows  full,  the 
coefficient  of  contraction  at 
the  outlet  equals  unity.  It 
has  been  found  experimentally 
that  the  coefficient  of  dis- 
charge, C,  and  therefore  the 
coefficient  of  velocity,  Cv,  for 
the  outlet  equals  approximately  0.82,  the  value  of  the  coefficient 
varying  slightly  with  the  head  and  diameter  of  tube.  Therefore 
from  formula  (14),  page  78 


FIG.  61. — Standard  short  tube. 


(32), 


STANDARD  SHORT  TUBE  87 

and  since  the  coefficient  of  contraction  is  unity, 

Q  =  CaV2^h  =  O.S2aV2ih  ......     (33) 

The  discharge  is  thus  about  one-third  greater  than  for  a  sharp- 
edged  orifice  of  the  same  diameter. 

To  investigate  conditions  at  the  contracted  portion  of  the  jet, 
Bernoulli's  equation  may  be  written  between  a  point  on  the 
water  surface,  where  the  velocity  is  considered  zero  and  the 
pressure  is  atmospheric,  and  a  point  m  in  the  contracted  portion 
of  the  jet.  Thus 


=      ++lost  head  .....     (34) 


Assuming  the  coefficient  of  contraction  at  m  to  be  0.62,  the  same 
as  for  a  sharp-edged  orifice  discharging  freely  into  the  air,  and 
writing  the  equation  of  continuity  between  m  and  n 


or 

vm=l.Glv  ........     (35) 

v  2 
The  head  lost  between  the  reservoir  and  m  (page  81)  is  0.04-^-. 

20 
Substituting  these  values,  equation  (34)  becomes 

,     .     .     (36) 
and  substituting  v  from  equation  (32)  and  reducing, 

^  =  34-0.8ft  ........     (37) 

There  exists,  therefore,  a  partial  vacuum  at  m  which  will  lift  a 
water  column  0.8ft  (Fig.  61),  the  pressure  being  O.Swft  less  than 
atmospheric  pressure.  This  has  been  confirmed  experimentally. 
Evidently  the  relation  does  not  hold  when  0.8ft  becomes  greater 
than  34  ft.,  or  when  the  head  becomes  greater  than  approxi- 

mately 42.5  ft.,  since  this  condition  gives  a  negative  value  to  — 

w 

in  equation  (37)  which  is  not  possible. 


FLOW  OF  WATpR  THROUGH  ORIFICES  AND  TUBES 

Jiead  in  /the  entire  length  of  a  standard  short  tube 
(see  Art.  54)  is  given  by  the  formula 


(22) 


v2 
'and  since  Cp  =  0.82,  the  formula  gives  h0  =  Q.5Q  —  . 


This  case  is  important  since  the  entrance  to  a  pipe,  where  the 
end  of  the  pipe  is  flush  with  a  vertical  wall,  is  usually  considered 
as  a  standard  short  tube  and  the  head  lost  at  entrance  to  the 
pipe  is  taken  as  the  head  lost  in  a  standard  short  tube  (see  page  156). 


FIG.  62. — Converging  tube  with 
sharp-cornered  entrance. 


FIG.  63. — Converging  tube  with 
rounded  entrance. 


59.  Converging  Tubes.  —  Converging  tubes  having  a  circular 
cross-section  are  frustums  of  cones  with  the  larger  end  adjacent 
to  the  reservoir.  They  may  have  a  sharp-cornered  entrance  as  in 
Fig.  62  or  a  rounded  entrance  as  in  Fig.  63.  The  jet  contracts 
slightly  at  a,  just  beyond  the  end  of  the  tube.  The  coefficient  of 
contraction,  Cc,  decreases  as  the  angle  of  convergence,  6,  increases; 
becoming  0.62  for  0=180°  when  the  tube  becomes  a  sharp-edged 
orifice.  The  coefficient  of  velocity,  Cv,  on  the  other  hand,  decreases 
as  6  decreases.  As  for  any  orifice 

(38) 


The  following  table  gives  coefficients  for  converging,  conical  tubes 
with  sharp-cornered  entrances,  interpolated  from  experiments  by 
,1'Aubuisson  and  Castel.  These  results  are  interesting  in  that 
Lhcy  show  the  general  laws  of  variation  of  coefficients  but,  on 


NOZZLES 


89 


account  of  the  small  models  used  in  the  experiments,  they  should 
not  be  taken  as  generally  applicable  to  all  tubes  of  this  type. 
COEFFICIENTS  FOR  CONICAL  CONVERGING  TUBES 


Coef- 

Angle  of  convergence,  6  (Fig.  62) 

ficient 

0° 

5° 

10° 

15° 

20° 

25° 

30° 

40° 

50° 

c. 

0.829 

0.911 

0.947 

0.965 

0.971 

0.973 

0.976 

0.981 

0.984 

Cc 

1.000 

.999 

.992 

.972 

.952 

.935 

.918 

.888 

.859 

C 

0.829 

.910 

.939 

.938 

.924 

.911 

.896 

.871 

.845 

The  coefficient  of  velocity  and  therefore  the  coefficient  of 
discharge  is  increased  by  rounding  the  entrance  (Fig.  63),  since 
this  reduces  the  head  lost  in  the  tube.  The  coefficient  of  con- 
traction will  not  be  materially  changed.  Exact  values  of  coeffi- 
cients will  depend  upon  the  extent  to  which  the  corner  is  rounded. 
Maximum  discharge  is  obtained  when  the  shape  of  the  entrance 
conforms  to  the  shape  of  the  contracted  jet. 

60.  Nozzles. — A  nozzle  is  a  converging  tube  attached  to  the 
end  of  a  pipe  or  hose.  The  nozzle  increases  the  velocity  of  the 
issuing  jet,  thus  increasing  the  range  of  distance  which  it  covers. 
Fig.  64  illustrates  two  types  of  nozzles  in  common  use.  Each  of 
these  has  a  cylindrical  tip  of  such  length  that  it  will  flow  full, 
thus  preventing  contraction  and  increasing  the  discharge.  The 
converging  part  of  the  tube  may 
be  the  frustum  of  a  cone  as  in 
Fig.  64  (a)  or  the  inside  may  be 
convex  as  in  (b).  Each  of  these 
shapes  gives  an  efficient  stream. 
The  following  mean  values  of  co- 
efficients of  discharge  for  smooth 
nozzles,  similar  to  Fig.  64  (a), 
having  a  diameter  at  the  base  of  1.55  in.,  have  been  determined 
from  experiments  by  Freeman.1 


FIG.  64. — Nozzles. 


Diameter  in  inches  

i 

4 

i 

1 

ii 

u 

If 

Coefficient  of  discharge.  . 

0.983 

0.982 

0.980 

0.976 

0.971 

0.959 

j 

1  JOHN  R.  FREEMAN:  Experiments  Relating  to  Hydraulics  of  Fire  Streams, 
Trans.  Amer.  Soc.  Civ.  Eng.,  vol.  21,  pp.  303-482  (1889). 


90       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 


Since  the  coefficient  of  contraction  is  unity,  the  coefficients  of 
discharge  given  above  are  also  the  coefficients  of  velocity. 

The  nozzle  being  a  form  of  tube  to  which  formula  (22),  page  81, 
applies,  the  head  lost  in  the  nozzle  is 


20' 


or  substituting  values  of  Cv  given  in  the  above  table, 

v2 


h0=  (0.04  to  0.09)^, 


(22) 


(39) 


in  which  v  is  the  mean  velocity  at  the  outlet  of  the  nozzle. 

Expressed  as  a  function  of  the  velocity,  vi,  in  the  hose  or  pipe 
having  a  diameter  D,  the  diameter  of  the  nozzle  being  d, 


.....     (40) 


Bernoulli's  equation,  for  a  horizontal  nozzle,  may  be  written  between 
a  point  at  entrance  to  the  nozzle  and  a  point  in  the  jet  as  follows: 


head> 


in  which  p\  is  the  gage  pressure  at  the  entrance,  v\  is  the  velocity 

at  entrance  and  v  is  the 
velocity  in  the  jet.  From 
this  equation  the  pressure 
at  the  base  of  the  nozzle, 
pi,  may  be  determined  if 
the  discharge  is  known  or 
the  discharge  may  be  de- 
termined if  pi  is  known. 

61.  Diverging  Tubes.— 
Fig.  65  represents  a   coni- 
cal diverging  tube,   having 
FIG.  65.— Diverging  tube.  rounded    entrance   corners, 

so     that     all     changes     in 

velocity  occur  gradually.  Such  a  tube,  provided  the  angle  of 
flare  is  not  too  great  nor  the  tube  too  long,  will  flow  full.  The 
theoretical  velocity,  vtj  at  the  outlet  of  the  tube,  obtained  in  the 
same  manner  as  for  an  orifice  (Arts.  51  and  52),  is 

(9) 


BORDA'S   MOUTHPIECE  91 

the  actual  mean  velocity  being 

v  =  C,V2gh>        (14) 

where  Cv  is  the  coefficient  of  velocity  at  the  outlet. 

Experiments  indicate  that  even  under  favorable  conditions 
the  value  of  Cv  is  small.  Venturi  and  Eytelwein,  experimenting 
with  a  tube  8  in.  long,  1  in.  in  diameter  at  the  throat  and  1.8  in. 
in  diameter  at  the  outer  end,  obtained  results  which  give  a  value 
of  Cv  of  about  0.46.  The  lost  head  (formula  (22),  Art.  54)  was, 
therefore,  approximately  0.79/i. 

Even  with  this  large  loss  of  head  the  discharge  through  the 
tube  was  about  two  and  one-half  times  the  discharge  from  a  sharp- 
edged  orifice  having  the  same  diameter  as  the  throat  of  the  tube. 

The  greater  portion  of  the  loss  of  head  occurs  between  the 
throat  and  outlet  of  the  tube  where  the  stream  is  expanding  and 
thus  has  a  tendency  to  break  up  in  eddies  with  a  waste  of  energy. 
Experiments  by  Venturi  indicate  that  an  included  angle,  6,  of 
about  5°  and  a  length  of  tube  about  nine  times  its  least  diameter 
give  the  most  efficient  discharge.  A  diverging  tube,  such  as  that 
shown  in  Fig.  65,  is  commonly  called  a  Venturi  tube. 

The  pressure  head  at  the  throat  is  evidently  less  than  atmos- 
pheric pressure.  This  may  be  shown  by  writing  Bernoulli's 
equation  between  ra  and  n.  When  the  throat  is  so  small  that 
Bernoulli's  equation  gives  a  negative  absolute  pressure  at  m,  for- 
mula (14)  no  longer  holds.  The  conditions  are  similar  to  those 
already  described  for  a  standard  short  tube,  Art.  58. 

62.  Borda's  Mouthpiece. — Since  the  contraction  of  a  jet  issuing 
from  an  orifice  is  caused  by  the  water  entering  the  orifice  from 
various  directions  inclined  to  the  axis  of  the  orifice,  it  follows 
that  the  greater  the  angle  between  the  extreme  directions  the 
greater  will  be  the  contraction  of  the  jet.  The  extreme  case 
occurs  in  Borda's  mouthpiece  (Fig.  66),  where  the  water 
approaches  the  orifice  from  all  directions.  This  mouthpiece 
consists  of  a  thin  tube  projecting  into  the  reservoir  about  one 
diameter.  The  proportions  are  such  that  the  jet  springs  clear 
of  the  walls  of  the  tube.  Borda's  mouthpiece  is  of  interest  because 
it  is  possible  to  obtain  its  coefficient  of  contraction  by  rational 
methods. 

The  cross-sectional  area  of  the  jet  at  the  vena  contracta,  mn, 


92       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 


is  a'  and  the  velocity  at  this  section  is  v.     If  a  is  the  area  of  the 

opening,  the  coefficient  of  contraction  Cc  =  a'/a. 

The  size  of  the  reservoir 
is  assumed  to  be  so  large  in 
comparison  with  the  area  of 
the  orifice  that  the  velocity 
of  the  water  within  the 
reservoir  may  be  neglected 
and  that  the  pressure  on  the 
walls  will,  therefore,  follow 
the  laws  of  hydrostatics. 
Excepting  the  pressure  act- 
ing on  the  horizontal  pro- 
jection de  of  the  mouthpiece 
on  the  opposite  wall,  the 
horizontal  pressures  on  the 
walls  will  balance  each 
other.  The  total  pressure 
on  de  is  wah,  which  is  also 


FIG.  66. — Borda's  mouthpiece. 


the  resultant  horizontal  accelerating  force   acting  on  the  water 
entering  the  mouthpiece. 

Consider  the  mass  of  water  xymn  to  move  to  the  position  x'y' 
m'n'  in  t  seconds.  The  change  in  the  momentum  of  the  mass  con- 
sidered is  the  difference  in  the  momentum  of  the  mass  xx'yy'  and 
mm'nn'.  But  the  momentum  of  xx'yy'  is  entirely  vertical,  there- 
fore the  change  in  momentum  in  a  horizontal  direction  is  equal  to 
the  momentum  of  mm'nn',  which  is  produced  by  the  action  of  the 
force  wah. 

rp,  ,.         ,     ,   .     a'vtw        ,   .  .     a'v2tw 

The  mass  of  mmnn   is  —       and  its  momentum  is   -      — . 

g.  g 

The  impulse  of  the  force  wah  is  waht.     Equating    impulse  and 

change  of  momentum, 

a'v2tw 
what  = 


therefore, 


and  since 


g 


a'  _gh 


v  = 
a/_  = 
a 


2C2' 


RE-ENTRANT  TUBES 


93 


Therefore,  assuming  the  coefficient  of  velocity  to  be  unity,  the 
coefficient  of  contraction  is  theoretically  0.5,  or  calling  the  coeffi- 
cient of  velocity  0.98,  the  same  as  for  a  sharp-edged  orifice,  the 
coefficient  of  contraction  is  0.52.  This  value  has  been  verified 
approximately  by  experiments. 

63.  Re-entrant  Tubes. — Tubes,  having  their  ends  project  into 
a  reservoir  (Fig.  67),  and  having  a  length  of  about  2^  diameters, 
are  called  re-entrant  or  inward-projecting  tubes.  The  action  of 
water  in  such  tubes  is  similar  to  that  in  standard  short  tubes 
(Art.  58),  except  that  the  contraction  of  the  jet  near  the  entrance 
is  greater,  At  the  discharge  end  the  tube  flows  full  and  the 


FIG.  67. — Re-entrant  tube. 


FIG.  68. — Submerged  orifice. 


coefficient  of  velocity  therefore  equals  the  coefficient  of  discharge. 
Thus  Cc=l  and  CV  =  C. 

From  experiments  Cp  =  0.75.     The  head  lost,  from  equation 

(22)  (Art.  54)  is,  therefore,  hQ  =  Q.7S  ~. 

29 

This  case  is  important  since  the  entrance  to  a  pipe,  which 
projects  into  a  body  of  water,  may  be  considered  as  a  re-entrant 
tube  and  the  head  lost  at  entrance  GO  the  pipe  is  taken  as  the  head 
lost  in  a  re-entrant  tube  (Art.  102). 

64.  Submerged  Orifice. — An  orifice  discharging  wholly  under 
water  (Fig.  68)  is  called  a  submerged  orifice.  The  assumption  is 
usually  made  that  every  filament  of  water  passing  through  the 
orifice  is  being  acted  upon  by  a  head,  hi-h2  =  h,  the  difference 
in  elevation  of  water  surfaces.  Based  upon  this  assumption 


94       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 


and  using  the  same  nomenclature  as  for  orifices  with  free  dis- 
charge 

(9) 


and 


(14) 
(18) 


Coefficients  of  discharge  for  sharp-edged,  submerged  orifices 
are  very  nearly  the  same  as  for  similar  orifices  discharging  into  the 
air. 

The  assumption  that  h,2  is  the  pressure  head  on  the  center  of 
the  orifice  at  its  lower  side  is  not  strictly  true  unless  all  of  the 
velocity  head,  due  to  the  velocity  of  the  water  leaving  the  orifice, 
is  lost  in  friction  and  turbulence  as  the  velocity  is  reduced  to  zero. 
It  has  been  .shown  experimentally  that  less  than  90  per  cent  of 
this  velocity  head  may  be  lost.  Assuming  a  loss  of  90  per  cent, 

v2 
the  pressure  head  at  the  center  of  the  orifice  is  h2  —  0. 12  5-.     The 

effect  of  this  condition  on  the  discharge  may  be  investigated  by 
writing  Bernoulli's  equation  between  m  and  n  and  n  and  p  (Fig. 
68).  This  matter  is  not  of  great  importance  in  connection  with 
submerged  orifices,  since  the  discrepancy  resulting  from  the  use  of 
formula  (18)  is  relatively  small  and  the  coefficient  of  discharge 
which  is  determined  from  experiments  eliminates  this  source  of 
error. 

The  loss  of  head  sustained  at  the  outlet  of  a  pipe  discharging 
into  a  body  of  still  water  is  discussed  in  Art.  102.  The  conditions 

of  discharge  in  this  case  are 
practically  identical  with 
those  of  the  submerged  orifice 
discussed  above. 

65.  Partially  Submerged 
Orifices. — Fig.  69  represents 
a  rectangular  orifice,  the  bot- 
tom of  which  is  submerged  to 
a  depth  D.  The  upper  and 
lower  edges  of  the  orifice  are, 
respectively,  hi  and  h2  below  the  upper  water  surface.  Z  is  the 
difference  in  elevation  of  water  surfaces.  L  is  the  length  of  the 


FIG.  69. — Partially  submerged  orifice. 


GATES  95 

orifice.  The  total  discharge  through  the  orifice  is  evidently  the 
combined  discharge  of  the  upper  portion  of  the  orifice  discharging 
into  the  air  and  the  lower  portion  discharging  as  a  submerged 
orifice. 

The  theoretical  formula  for  discharge  from  this  orifice  is  inter- 
esting because  of  its  relation  to  the  submerged  weir  (Art.  79). 
Let  Qi  and  Q2  be,  respectively,  the  discharges  from  the  free  and 
submerged  portions  of  the  orifice.  Then  from  Art.  56,  if  C"  is 
the  coefficient  of  discharge  for  the  upper  portion 

h,*),    .........     (42) 


and  by  Art.  64,  C"  being  the  coefficient  of  discharge  for  the  lower 
portion 

Z),      .........     (43) 


and  the  total  discharge,  Q,  for  the  orifice  is 

Q  =  Qi  +Q2  =  LVfy[%C'(Z3A  -  hi*}+C"Vz(h2  -  Z)],  (44) 
or  since  1i2  —  Z  =  D, 

~]  .....     (45) 


Since  the  coefficient  of  discharge  for  an  orifice  with  free  discharge 
is  very  nearly  equal  to  the  coefficient  for  a  submerged  orifice  the 
equation  may  be  put  in  the  approximately  equivalent  form 

(46) 


If  /ii  =  0  the  orifice  is  a  submerged  weir  and  equation   (45) 
becomes 

(47) 


The  submerged  weir  is  discussed  in  Arts.  79  and  80. 

66.  Gates.  —  As  used  in  engineering  practice  gates  are  forms 
of  orifices.  They  may  discharge  freely  into  the  air  or  be  partially 
or  wholly  submerged.  Though  the  principles  underlying  the  dis- 
charge through  orifices  have  been  discussed  in  the  preceding  pages 
they  cannot  be  applied  accurately  to  gates  because  of  the  fact  that 
gates  do  not  ordinarily  conform  to  the  regular  sections  for  which 
coefficients  are  directly  available. 


96       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 

Fig.  70  illustrates  a  cross-section  of  a  head  gate  such  as  is 
commonly  used  in  diverting  water  from  a  river  into  a  canal.  A 
curtain  wall  extends  between  two  piers,  having  grooves  in  which 
the  gate  slides.  The  bottom  of  the  opening  is  flush  with  the  floor 
of  the  structure.  Such  an  opening  has  suppressed  contraction 
at  the  bottom,  nearly  complete  contraction  at  the  top  and  par- 
tially suppressed  contractions  at  the  sides.  Other  equally  com- 
plex conditions  arise.  The  selection  of  coefficients  for  gates  is 
therefore  a  matter  requiring  mature  judgment  and  an  intelligent 
use  of  the  few  available  experimental  data.  Even  the  most 


FIG.  70.— Headgate. 


FIG.  71. — Discharge  under 
falling  head. 


experienced  engineers  may  expect  errors  of  at  least  10  per  cent 
in  the  coefficients  which  they  select  and  to  provide  for  this  uncer- 
tainty ample  allowance  should  be  made  in  designs. 

67.  Discharge  under  Falling  Head. — A  vessel  is  filled  with 
water  to  a  depth  hi  (Fig.  71).  It  is  desired  to  determine  the  time 
required  to  lower  the  water  surface  to  a  depth  h,2  through  a  given 
orifice.  A  is  the  area  of  the  water  surface  when  the  depth  of  water 
is  y  and  a  is  the  area  of  the  orifice.  The  rate  of  discharge  at  any 
instant  when  the  head  is  y,  the  coefficient  of  discharge  being  C,  is 


CaV2gy, 


DISCHARGE   UNDER   FALLING   HEAD 


97 


and  in  the  infinitesimal  time,  dt,  the  corresponding  volume  of 
water  which  flows  out  is 


In  the  same  infinitesimal  time  the  head  will  drop  dy  and  the 
volume  of  water  discharged  will  be 


Equating  the  values  of  dV 


=  CaV2gydt 


or 


dt= 

Ca\/2gy 


(48) 


From  this  expression,  by  integrating  with  respect  to  y  between 
the  limits  hi  and  /i2,  the  time  required  to  lower  the  water  surface 
the  amount  (hi  —  fo)  may  be  determined  or  the  time  of  emptying 
the  vessel  may  be  obtained  by  placing  /i2  =  0,  provided  A  can  be 
expressed  in  terms  of  y.  For  a  cylinder  or  prism  the  cross-sectional 
area,  A,  is  constant  and  the  formula  after  integration  becomes 


t  = 


2A 


CaV2g 


(49) 


The  above  formulas  apply  also  to  vertical  or  inclined  orifices, 
provided  the  water  surface  does  not  fall  below  the  top  of  the 
orifice.  The  heads  hi  and  h2  are  then  measured  to  the  center  of 
the  orifice.  The  time  required  to  completely  empty  a  vessel 
evidently  can  be  determined  only  in  the  case  of  a  horizontal 
orifice. 

Example. — Two  chambers,  1  and  2  (Fig.  72),  with  vertical 
sides,  each  chamber  being  8  ft. 
wide,  are  separated  by  a  par- 
tition. Chamber  1  is  25  ft. 
long  and  chamber  2  is  10  ft. 
long.  At  the  bottom  of  the 
partition  is  an  orifice  1  ft.  by 
2  ft.  The  orifice  is  at  all  times 
submerged.  The  coefficient  of 
discharge  is  0.85.  At  a  certain  instant  the  water  surface  is  10 
ft.  higher  in  chamber  1  than  in  chamber  2.  After  what  interval 


10' 


FIG.  72. 


98       FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 

of  time  will  the  water  surfaces   in  the  two  chambers  be  at  the 
same  elevation? 

Solution.  —  Let  y  be  the  difference  in  elevation  of  water  surfaces 
at  any  instant  and  dy  be  the  change  in  the  difference  in  elevation 
of  water  surfaces  in  time  dt.  The  amount  of  water  flowing  into 
chamber  2  in  time  dt  will  be 

dV  =  CaVtyy  dt  =  0  .  85  X  2  X  8  .  Q2Vydt  =  13  .  6  Vydt. 

Also  in  the  same  interval  of  time   the  head  will  drop  ^|  dy  in 
chamber  1  and  rise  f  f  dy  in  chamber  2.     Then 

„,     10X25X8,      2000 
dV= 
Equating  values  of  dV 


or 

Civ  —  ,  _         • 

Vy 

Integrating  between  the  limits  10  and  0  and  reducing, 

2  =  26.5  seconds. 

PROBLEMS 

1.  A  sharp-edged  orifice,  2  in.  in  diameter,  in  the  vertical  side  of  a  large 
tank,  discharges  under  a  head  of  16  ft.     If  Cc  is  0.62  and  Cv  is  0.98  determine 
the  diameter  and  velocity  of  the  jet  at  the  vena  contracta  and  the  discharge 
in  cubic  feet  per  second. 

2.  In  Problem  1  how  far  from  the  vertical  plane  containing  the  orifice 
will  the  jet  strike  a  horizontal  plane  which  is  6  ft.  below  the  center  of  the 
orifice? 

3.  A  sharp-edged  orifice,  3  in.  in  diameter,  lies  in  a  horizontal  plane,  the 
jet  being  directed  upward.     If  the  jet  rises  to  a  height  of  21.2  ft.  and  the 
coefficient  of  velocity  is  0.98,  what  is  the  depth  of  the  orifice  below  the  water 
surface,  neglecting  air  friction.     The  pressure  in  the  jet  and  on  the  surface 
of  the  reservoir  is  atmospheric. 

4.  In  Problem  3,  if  Cc  =  0.62,  what  is  the  diameter  of  the  jet  16  ft.  above 
the  orifice? 

6.  If  the  orifice  shown  in  Fig.  49,  page  73,  has  a  diameter  of  2  in.  and  the 
diameter  of  the  vena  contracta  is  1.6  in.  determine  the  discharge  if  A  =  3.6  ft., 
F  =  0,  pA  =9  .7  Ibs.  per  square  inch,  p^  =  1.3  Ibs.  per  square  inch  and  the 
head  lost  is  0.8  ft. 

6.  A  sharp-edged  orifice,  4  in.  in  diameter,  in  the  vertical  wall  of  a  tank 
discharges  under  a  constant  head  of  4  ft.  The  volume  of  water  discharged 


PROBLEMS  99 

in  2  minutes  weighs  6352  Ibs.  At  a  point  2.57  ft.  below  the  orifice  the  center 
of  the  jet  is  6.28  ft.  distant  horizontally  from  the  orifice.  Determine  Cc,  Cv 
andC. 

7.  Determine  the  theoretical  discharge  (neglecting  velocity  of  approach) 
from  a  vertical  rectangular  orifice  3  ft.  long  and  1  ft.  high,  the  head  on  the 
top  of  the  orifice  being  2  ft. 

8.  A  standard  short  tube,  4  in.  in  diameter,  discharges  under  a  head  of 
20  ft.     What  is  the  discharge  in  cubic  feet  per  second?     In  gallons  per  day? 

9.  If  a  s-in.  hole  is  tapped  into  the  standard  short  tube,  referred  to  hi 
Problem  8,  at  a  point  2  in.  from  the  entrance,  determine  the  discharge  through 
the  tube,  assuming  the  friction  losses  to  remain  the  same. 

10.  If  the  upper  end  of  a  piezometer  tube  is  connected  with  the  ^-in.  hole 
referred  to  in  Problem  9  and  the  lower  end  is  submerged  in  a  pan  of  mercury, 
to  what  height  will  the  mercury  rise  in  the  tube? 

11.  A  Borda's  mouthpiece  6  in.  in  diameter  discharges  under  a  head  of 
10  ft.     What  is  the  discharge  in  cubic  feet  per  second?     What  is  the  diameter 
of  the  jet  at  the  vena  contracta? 

12.  Water  is  discharging  through  a  gate  18  in.  square.     On  the  upstream 
side  the  water  surface  is  5  ft.  above  the  top  of  the  gate  and  on  the  down- 
stream side  it  is  2  ft.  above.     If  the  coefficient  of  discharge  is  0.82,  what  is  the 
discharge  in  cubic  feet  per  second? 

13.  A  canal  carrying  40  cu.  ft.  per  second  has  a  depth  of  water  of  3  ft. 
A  structure  is  built  across  the  canal  containing  a  gate  2  ft.  square,  the  bottom 
of  the  gate  being  set  flush  with  the  bottom  of  the  canal.     If  the  coefficient  of 
discharge  is  0.85,  what  will  be  the  depth  of  water  on  the  upstream  side  of  the 
gate? 

14.  If,  in  Problem  13,  the  gate  has  a  width  of  3  ft.  and  it  is  desired  to 
increase  the  depth  of  water  above  the  structure  to  4  ft.,   what  should  be 
the  height  of  the  gate,  all  other  conditions  remaining  the  same? 

15.  A  3-in.  fire  hose  discharges  water  through  a  nozzle  having  a  diameter 
at  the  tip  of  1  in.     If  there  is  no  contraction  of  the  jet  and  (7^  =  0.97,  the 
gage  pressure  at  the  base  of  the  nozzle  being  60  Ibs.  per  square  inch,  what  is  the 
discharge  in  gallons  per  minute? 

16.  In  Problem  15  to  what  vertical  height  can  the  stream  be  thrown, 
neglecting  air  friction? 

17.  In  Problem  15,  if  it  is  desired  to  throw  a  stream  to  a  vertical  height  of 
100  ft.,  what  must  be  the  pressure  at  the  base  of  the  nozzle? 

18.  In  Problem  15  what  is  the  maximum  horizontal  range  (in  the  plane  of 
the  nozzle)  to  which  the  stream  can  be  thrown? 

19.  A  fire  pump  delivers  water  through  a  6-in.  main  to  a  hydrant  to  which 
is  connected  a  3-in.  hose,  terminating  in  a  1-in.  nozzle.     The  nozzle,  for  which 
Cc  =  l  and  Cp  =  0.97,  is  10  ft.  above  the  hydrant  and  the  hydrant  is  50  ft. 
above  the  pump.     What  gage  pressure  at  the  pump  is  necessary  to  throw 
a  stream  80  ft.  vertically  above  the  nozzle? 

20.  A  cylindrical  vessel  4  ft.  in  diameter  and  6  ft.  high  has  a  sharp-edged 
circular  orifice  2  in.  in  diameter  in  the  bottom.     If  the  vessel  is  filled  with 
water  how  long  will  it  take  to  lower  the  water  surface  4  ft.? 

21.  A  tank,  which  is  the  frustum  of  a  cone  having  its  bases  horizontal 


100     FLOW  OF  WATER  THROUGH  ORIFICES  AND  TUBES 

and  axis  vertical,  is  10  ft.  high  and  filled  with  water.  It  has  a  diameter  of 
8  ft.  at  the  top  and  3  ft.  at  the  bottom.  What  is  the  time  required  to  empty 
the  tank  through  a  sharp-edged  orifice  3  in.  square? 

22.  A  hemispherical  shell,  with  base  horizontal  and  uppermost,  is  filled 
with  water.     If  the  radius  is  8  ft.  determine,  the  time  required  to  empty 
through  a  sharp-edged  orifice  6  in.  in  diameter  located  at  the  lowest  point. 

23.  A  tank  12  ft.  long  has  its  ends  vertical,  top  and  bottom  horizontal, 
and  is  6  ft.  high.     The  top  and  bottom  are  rectangular,  having  widths  of  8  ft. 
and  5  ft.,  respectively.     A  standard  short  tube,  4  in.  in  diameter,  is  located 
in  one  end  near  the  bottom.     If  at  the  beginning  the  tank  is  full,  find  the 
time  necessary  to  lower  the  water  surface  4  ft. 

24.  In  the  tank  described  in  Problem  23  assume  that  there  is  a  vertical 
partition  parallel  with  the  ends  and  5  ft.  distant  from  one  end.     Near  the 
bottom  of  this  partition  there  is  a  circular,  sharp-edged  orifice  4  in.  in  diameter. 
If  at  the  beginning  the  larger  chamber  is  filled  and  the  smaller  chamber  con- 
tains water  having  a  depth  of  2  ft.,  find  the  time  required  for  the  water 
surfaces  to  come  to  the  same  level. 


CHAPTER  VIII 


FLOW  OF  WATER  OVER  WEIRS 

68.  Description  and  Definitions. — A  weir  may  be  described  as 
any  notch  of  regular  form  through  which  water  flows.  This  notch 
may  be  in  the  side  of  a  tank,  reservoir  or  channel  or  it  may  be  an 
overflow  dam  with  retaining  walls  at  its  ends.  In  general  any 
obstruction,  having  an  approximately  uniform  cross-section,  placed 
in  a  channel  so  that  water  must  flow  over  it  is  a  weir. 

The  edge  or  surface  over  which  the  water  flows  is  called  the 
crest  of  the  weir.  The  overfalling  sheet  of  water  has  been  termed 
the  nappe. 

Weirs  may  be  classified  in  two  ways,  (a)  with  reference  to  the 
shape  of  the  notch  and  (6)  with  reference  to  the  cross-sectional 
form  of  the  crest. 

Rectangular  weirs — that  is,  weirs  having  a  level  crest  and 
vertical  sides,  are  the  most  generally  used.  Other  weirs  in  more 
or  less  common  use,  named  from  the  shape  of  the  notch  or  opening, 
are  triangular  weirs,  trapezoidal  weirs  and  parabolic  weirs. 

Weir  crests  are  constructed  of  many  cross-sectional  forms,  but 
they  all  come  under  one  of  the  general  headings,  (a)  sharp-crested 
weirs,  which  are  used  primarily  for  the  measurement  of  flowing 

water  and  (6)  weirs  not  sharp    

crested    which   are   used  pri-  :  5^^? 

marily  as  a  part  of  hydraulic 

structures. 

A  sharp-crested  weir  is  a 
weir  with  a  sharp  upstream 
edge  so  formed  that  water  in 
passing  touches  only  this  edge. 
The  nappe  from  such  a  weir 


FIG.  73. — Sharp-crested  weir. 


is  contracted  at  its  under  side  in  the  same  way  that  the  jet  from 
a  sharp-edged  orifice  is  contracted.  This  is  called  crest  contrac- 
tion. If  the  sides  of  the  notch  also  have  sharp  upstream  edges  so 

101 


OF  WATER  OVER  WEIRS 


that  the  nappe  is  contracted  in  width  the  weir  is  said  to  have  end 
contractions.  The  nappe  from  a  weir  having  a  length  equal  to  the 
width  of  the  channel  suffers  no  contraction  in  width  and  such  a 
weir  is  said  to  have  end  contractions  suppressed.  Fig.  73  is  a 
cross-section  of  a  sharp-crested  weir  which  illustrates  crest  con- 
traction. Figs.  74  and  75  are  views  of  weirs  with  end  contrac- 
tions. Fig,  76  shows  a  weir  with  end  contractions  suppressed. 


FIG.  74. — WTeir  with  end  contractions. 

There  is  a  downward  curvature  to  the  water  surface  near  the 
weir  crest  (Fig.  73).  This  is  called  the  surface  contraction.  The 
head,  H  (Fig.  73),  is  the  vertical  distance  from  the  water  surface, 
back  of  the  effects  of  surface  contraction,  to  the  crest  of  the  weir. 
The  curvature  of  the  water  surface  is  not  perceptible  beyond  a 
distance  of  about  2H  upstream  from  the  weir.  The  head  is 
usually  measured  at  distances  of  6  to  16  ft.  upstream  from  the  weir. 


DESCRIPTION   AND   DEFINITIONS 


103 


The  vertical  contraction  of  the  nappe  includes  both  the  surface 
contraction  and  the  crest  contraction.  The  section  where  the 
effects  of  crest  contraction  disappear,  corresponding  to  the  vena 
contracta  of  the  jet,  will  be  referred  to  as  the  contracted  section  of 
the  nappe. 

Incomplete  contraction  of  the  nappe  occurs  when  the  crest  of  a 
weir  is  so  near  the  bottom,  or  the  ends  of  a  weir  with  end  con- 
tractions are  so  near  to  the  sides  of  the  channel,  as  to  interfere  with 
the  approach  of  the  water  filaments  in  directions  parallel  to  the 
face  of  the  weir.  The  conditions  are  similar  to  those  causing  par- 


FIG.  75. — Weir  with  end  contractions. 

tial  suppression  of  the  contraction  of  the  jet  issuing  from  an 
orifice,  discussed  in  Art.  57. 

Weirs  not  sharp  crested  are  constructed  in  a  wide  variety  of 
cross-sectional  forms  as  is  exemplified  in  the  many  shapes  of  over- 
flow dams  now  in  existence.  Such  weirs  have  surface  contraction 
similar  to  sharp-crested  weirs,  but  conditions  at  the  crest  are 
different  and  vary  with  the  sectional  form  (see  Figs.  84  to  87). 
A  variety  of  cross-sections  of  weirs  of  this  class  are  shown  in 
Fig.  88. 

The  term  velocity  of  approach,  as  used  in  connection  with  weirs, 
means  the  mean  velocity  in  the  channel  just  upstream  from  the 
weir.  The  portion  of  the  channel  near  where  the  head  is  measured 
is  designated  the  channel  of  approach.  The  height  of  a  weir,  P 
(Fig.  73),  is  the  vertical  distance  of  the  crest  above  the  bottom 
of  the  channel  of  approach. 


104 


FLOW  OF  WATER  OVER  WEIRS 


69.  Velocity  at  any  Depth. — Consider  water  to  be  discharging 
over  the  weir  crest  A  (Fig.  77).  In  the  derivation  of  the  funda- 
mental formula  it  will  be  assumed  that  the  water  flows  without 
friction  and  also  that  there  is  no  contraction  of  the  nappe  and 
therefore  no  pressure  within  the  nappe.  In  order  to  write  a  gen- 
eral expression  applicable  to  all  filaments  it  will  be  necessary  to 
make  the  further  assumption  that  all  of  the  water  particles  in 


FIG.  76. — Weir  with  end  contractions  suppressed. 

a  cross-section  of  the  channel  of  approach  flow  with  the  same 
velocity. 

From  the  nature  of  the  discharge  over  weirs  it  is  evident  that 
the  water  surface  in  the  channel  and  the  nappe  must  be  subjected 
to  the  same  pressure  from  surrounding  gases,  which  is  usually 
atmospheric  pressure.  All  pressures  excepting  those  resulting 
from  the  weight  of  the  water  may  therefore  be  neglected. 

The  flowing  water  may  be  considered  to  be  made  up  of  filaments 
of  which  mn  is  one,  m  being  a  point  in  the  channel  of  approach 


THEORETICAL  FORMULAS   FOR  DISCHARGE  105 

and  n  a  point  in  the  nappe,  in  the  plane  of  the  weir.  The  filament 
passes  over  the  weir  at  a  distance  y  below  the  surface  of  the  water. 
The  point  m  is  a  distance  hm  below  the  water  surface  and  a  distance 
z  below  n.  vm  is  the  velocity  at  m  and  vn  is  the  velocity  at  n. 
Bernoulli's  equation  may  be  written  between  the  points  m  and  n 
as  follows: 


and  since  hm—z  = 
and 


(3) 


These   formulas   express  the  theoretical  relation  between  depth 
and  velocity  for  any  point  in  the  plane  of  the  weir. 

Introducing  the  assumption  that  all  of  the  water  particles  in  a 
cross-section  of  the  channel  of  approach  flow  with  the  same 
velocity,  V;  vm  in  formulas  (2)  and  (3)  may  be  replaced  by  V, 
which  gives 


and 

(5) 

If  the  cross-sectional  area  of  the  channel  of  approach  is  very 
much  larger  than  the  area  of  the  notch,  the  velocity  of  approach 
is  small  and  V  may  be  called  zero.  The  depth,  y,  at  which  the 
velocity  vn  occurs  is  then  from  formula  (4) 


and  the  theoretical  velocity  at  a  depth  y,  from  formula  (5)  is 

vn  =  V2^.     ,     .     .     .     .     ...     .     .     (7) 

70.  Theoretical  Formulas  for  Discharge.  —  Referring  again  to 
Fig.  77,  let  an  origin  be  assumed  at  0,  a  distance  H  vertically 


106 


FLOW  OF  WATER  OVER  WEIRS 


above  the  weir  crest  A.     Formula  (5)  as  derived  in  the  preceding 
article  is 


This  is  also  the  equation  of  a  parabola  whose  axis  is  the  line  OA 

V2 
and  whose  intersection  with  the  axis  is  at  M ,  a  distance  —  above 

*9 
the  origin.  Assuming  the  curve  MN  to  be  the  graph  of  the 


FIG.  77. 


equation,  the  abscissa  at  any  depth,  y,  is  the  theoretical  velocity  at 
this  depth. 

Considering  a  unit  length  of  weir,  the  area  of  an  elementary 
strip  is  dy  and  the  theoretical  discharge  through  this  strip  is 


dQi=vndy 

Substituting  the  value  of  vn  from  equation  (5) 


(8) 


dQi  = 


and 


•5T   * 


THEORETICAL  FORMULA  FOR   MEAN   VELOCITY         107 
or 

-©*]•  •  •  •  w 

This  formula  expresses  the  theoretical  discharge  over  a  weir 
1  ft.  long,  assuming  uniform  velocity  in  a  cross-section  of  the 
channel  of  approach,  and  neglecting  the  effects  of  friction  and 
contraction  of  the  nappe.  It  evidently  is  also  the  area  of  the 
surface  OSNA  (Fig.  77). 

As  a  matter  of  convenience  the  symbol  h  may  be  substituted 

V2 
for  -^-.     Making  this  substitution  the  theoretical  discharge  for  a 

weir  of  length  L  becomes 

which  formula  may  be  transposed  to  the  form 

'  TA  *n 

.    .    (H) 

In  this  form  the  term  within  the  brackets  is  the  factor  which 
corrects  for  velocity  of  approach.  If  the  cross-sectional  area  of 
the  channel  of  approach  is  large  in  comparison  with  the  cross- 
sectional  area  of  the  nappe,  the  effect  of  velocity  of  approach  will 
not  be  appreciable  and  may  be  considered  to  be  zero.  The  above 
formulas  then  reduce  to 


(12) 

which  is  the  same  as  formula  (27)  (page  83). 

This  formula  may  also  be  derived  directly  from  Fig.  77. 
The  area  of  the  surface  A  OP  which  represents  the  discharge  over  a 
weir  1  ft.  long,  being  half  of  a  parabolic  segment,  is  equal  to  two- 
thirds  of  the  area  of  the  circumscribed  .rectangle  ORPA  or 
%HV2gH.  The  discharge  for  a  weir  of  length  L  is  therefore 
%L/V2gH%,  which  is  the  same  as  formula  (12). 

71.  Theoretical  Formula  for  Mean  Velocity. — Since  formula 
(9)  which  is  the  theoretical  formula  for  discharge  over  a  weir  1  ft. 
long  is  also  an  expression  for  the  area  of  the  surface  OSNA  and 
since  the  abscissas  to  this  curve  at  any  depth  are  the  velocities 
at  the  depth,  the  mean  of  the  abscissas  between  0  and  A  gives 
the  mean  velocity  of  the  water  discharging  over  the  weir.  The 


108  FLOW  OF  WATER  OVER  WEIRS 

mean  velocity  is  therefore  the  area  of  the  surface  OSNA  divided 
by  H,  and  the  expression  for  theoretical  mean  velocity  is  obtained 
by  dividing  formula  (9)  by  H,  which  gives,  after  substituting  h 


If  the  velocity  of  approach  is  considered  to  be  zero,  h  also 
becomes  zero  and  the  above  formula  reduces  to 


(14) 


Equating  the  right-hand  members  of  equations  (7)  and  (14)  gives 
the  theoretical  depth  at  which  the  mean  velocity  occurs,  or 

y  =  *H  .........     (15) 


72.  Weir  Coefficients.  —  The  assumptions  which  were  made  in 
the  derivation  of  formula  (10)  may  be  summarized  briefly  as  follows: 

(a)  All  water  particles  in  a  cross-section  of  the  channel  of 
approach  flow  with  the  same  velocity. 

(6)  There  is  no  contraction  of  the  nappe. 

(c)  The  water  flows  without  friction. 

Since  these  conditions  do  not  in  reality  exist,  it  is  necessary 
to  modify  formula  (10)  and  the  formulas  derived  therefrom  to  make 
them  applicable  to  actual  conditions.  To  accomplish  this,  three 
empirical  coefficients  are  applied  to  the  formula,  there  being  one 
coefficient  to  correct  for  the  difference  between  assumed  conditions 
and  actual  conditions  for  each  of  the  above  assumptions.  The 
method  of  correcting  for  each  assumption  will  be  discussed  in  the 
order  given  above. 

(a)  Correction  for  non-uniformity  of  velocity  in  cross-section  of 
channel  of  approach.  The  velocity  in  any  cross-section  of  a  channel 
is  never  uniform.  As  a  result  of  the  combined  effects  of  friction, 
viscosity  and  surface  tension  (Arts.  7  and  110)  velocities  are 
lowest  near  the  sides  and  bottom  of  an  open  channel  and,  if  the 
channel  is  straight  and  uniform,  the  maximum  velocity  is  below 
the  surface  and  near  the  center  of  the  channel.  If  there  are  no 
obstructions,  velocities  in  a  vertical  line  (Art.  110)  vary  approxi- 
mately as  the  abscissas  to  a  parabola.  In  the  channel  of 
approach  where  a  weir  obstructs  the  flow,  the  law  of  distribution 


WEIR  COEFFICIENTS 


109 


of  velocities  is  not  well  understood  and  in  cases  where  these 
velocities  have  been  measured  they  have  been  found  to  vary  quite 
irregularly.  It  is  not  practicable  therefore  to  determine  by 
analysis  the  extent  to  which  discharges  over  a  weir  may  be  affected 
by  the  distribution  of  velocities  in  a  cross-section  of  the  channel  of 
approach,  but  the  general  effect  may  be  seen  by  studying  certain 
assumed  conditions. 

Let  the  curve  CMB  (Fig.  78)  represent  any  vertical  distribu- 
tion of  velocities  in  the  channel  of  approach  for  a  strip  of  water 


O' 


e=  3 

— 

<  ^-!->|  /                      ^^ 

d 

^dy   j    y 

^\ 

/I'M 

//'! 

V 

' 

A 

'/BX-X/     !F 

\ 

FIG.  78.  —  Velocities  in  channel  of  approach. 

1  ft.  wide.     The  velocity  at  any  distance,  y,  above  the  bottom  is  v, 
the  total  depth  being  d.     The  kinetic  energy  for  this  strip  of  water 


-*-     -* 


The  kinetic  energy  for  any  distribution  of  velocities  can  be  deter- 
mined from  this  formula  where  v  can  be  expressed  in  terms  of  y. 

Three  conditions  of  assumed  velocities  are  illustrated  in  Fig. 
78.  Uniform  velocities  are  indicated  by  the  vertica.1  line  EF. 
Velocities  decreasing  uniformly  downward  with  a  bottom  velocity 
of  zero  are  illustrated  by  the  line  DA.  The  line  CM  B  illustrates 
a  parabolic  distribution  of  velocities.  As  these  three  lines  are 
drawn,  the  mean  velocity,  V,  is  the  same  for  each  case. 

For  uniform  velocities,  v  in  equation  (16)  is  constant  and  equal 
to  V,  Substituting  this  value  and  integrating,  there  results 


KE 


wdV* 


(17) 


110  FLOW  OF  WATER  OVER  WEIRS 

For  uniformly  varying  velocities,  illustrated  by  the  line  DA 
(Fig.  78)  v=2V^.  Substituting  this  value  in  equation  (16)  and 
integrating 


which  shows  the  kinetic  energy  for  this  distribution  of  velocities 
to  be  twice  as  great  as  for  uniform  velocity. 

It  may  be  shown  also  by  writing  the  equation  of  a  curve 
similar  to  CMS,  v  being  expressed  as  a  function  of  y,  and  sub- 
stituted for  v  in  equation  (16),  that  the  kinetic  energy  for  this 
distribution  of  velocities  is  about  1.3  times  the  kinetic  energy  for 
uniformly  distributed  velocities. 

Similarly  for  any  variation  in  velocities  in  the  cross-section  of  a 
channel,  it  may  be  shown  that  the  water  contains  more  kinetic 
energy  if  the  velocity  is  non-uniform  than  if  it  is  uniform. 

In  general,  the  kinetic  energy  contained  in  the  water  in  the 
channel  of  approach  may  be  written 


(19) 


in  which  a  is  an  empirical  coefficient  always  greater  than  unity, 
and  since  velocity  head  is  the  kinetic  energy  contained  in  1  Ib.  of 

water  (Art.  43)  the  general  expression  for  velocity  head  due  to 

72 
velocity  of  approach  is  a  -g-  or  vh. 

This  expression  should,  therefore,  be  written  for  h  in  formula 
(13),  and  calling  v'  the  velocity  after  this  correction  has  been 
applied 


•  •  • 

(6)  Correction  for  contraction.  —  The  ratio  of  the  thickness  of 
the  nappe  at  its  contracted  section  to  the  head  on  the  weir  may  be 
called  the  coefficient  of  contraction,  Cc.  This  includes  only  vertical 
contraction;  a  separate  correction  being  required  for  weirs  with 
end  contractions  (see  Art.  73).  If  t  is  the  thickness  of  the  nappe 
and  H  the  head 

Ce  =        or    t  =  CcH 


WEIR  COEFFICIENTS  111 

and  if  v  is  the  actual  mean  velocity  in  the  contracted  section  of  the 
nappe,  L  being  the  length  of  the  weir,  the  discharge  over  the  weir  is 

Q  =  tLv  =  CcLHv.      ..'..,     .     .     .     (21) 

The  average  value  of  Ce  for  .sharp-crested  weirs  is  about  0.635. 

(c)  Correction  for  friction.  The  velocity  in  the  nappe  suffers  a 
a  retardation  by  reason  of  the  combined  effects  of  friction  and 
viscosity.  The  ratio  of  the  actual  mean  velocity,  vy  to  the  velocity 
*/,  which  would  exist  without  friction,  is  called  the  coefficient  of 
velocity,  Designating  the  coefficient  of  velocity  by  the  symbol  CVj 

Cv  =  —f    or    v  =  Cvi/9 

substituting  this  value  of  v  in  (21) 

Q  =  CcCvLHv'  .........     (22) 

The  average  value  of  CP  for  a  sharp-crested  weir  is  probably  about 
0.98,  the  same  as  for  a  sharp-edged  orifice. 

Substituting  the  value  of  v'  given  in  formula  (20),  the  formula 
for  discharge  over  a  weir  with  the  three  corrective  coefficients 
becomes 

.     .     (23) 

It  is  usual  to  combine  §  V2gCcCv  into  a  single  coefficient,  C,  called 
the  weir  coefficient,  then 

(24) 


If  Co  =  0.635  and  C,  =  0.98,  the  values  given  above,  (7  =  3.33, 
which  is  an  average  value  of  this  coefficient.  It  is  the  value  adopted 
by  Francis  as  a  result  of  his  experiments  on  sharp-crested  weirs. 

Later  experiments  by  Fteley  and  Stearns,  and  Bazin  con- 
sidered in  connection  with  the  Francis  experiments,  show  quite 
conclusively  that  C  is  not  a  constant.  Its  value  appears  to  be  repre- 
sented quite  closely  by  the  expression 

3.34 

~  ^0.03' 


112  FLOW  OF  WATER  OVER  WEIRS 

An  investigation  by  Bazin  gave  the  following  value  of  C: 


With  C  substituted,  formula  (23)  becomes 


The  expression  within  the  brackets  is  the  correction  for  velocity 
of  approach.  When  the  velocity  of  approach  is  so  small  that  the 
head,  h,  due  to  this  velocity  may  be  considered  zero,  formula  (25) 
becomes 

(26) 


Formula  (25)  includes  coefficients  which  correct  for  all  of  the 
assumptions  which  were  made  in  deriving  the  theoretical  formula 
(10)  for  weirs  with  end  contractions  suppressed. 

Formula  (25)  is  often  written  in  the  equivalent  form 

)*-(<zh)K]  ......     (27) 


73.  Weirs  with  End  Contractions.  —  The  weir  coefficient,  C, 
does  not  include  end  contractions.     A  separate  correction  must 
therefore  be  applied  to  the  above  formulas  to  make  them  applicable 
to  weirs   with  end  contractions.     End   contractions   reduce  the 
effective  length  of   a  weir.     Francis    determined    from    his   own 
experiments  that  the  effective  weir  length  is  reduced  an  amount 
equal  to  0.  \H  by  each  contraction.    If  L  is  the  effective  length  of 
the  weir,  L'  the  measured  length  and  N  the  number  of  contrac- 
tions, from  Francis'  determination  (see  Fig.  75) 

L  =  L'-O.INH.       ......     (28) 

For  two  end  contractions  N  =  2.     If  contraction  is  suppressed  at 
one  end  TV  =1. 

Some  of  the  later  experiments  do  not  substantiate  the  results 
of  Francis,  but  no  general  formula  better  than  the  above  has  been 
suggested.  On  account  of  uncertainty  regarding  the  best  method 
of  correcting  for  end  contractions,  where  they  can  be  properly 
used,  weirs  with  end  contractions  suppressed  are  preferable. 

74.  Modifications    of   Fundamental   Formula.  —  In   the   form 
given,  formula  (25)  or  its  equivalent  (27)  is  cumbersome  and  not 


ALGEBRAIC  TRANSFORMATION  OF  FORMULA          113 

convenient  to  use  as  a  base  formula.  It  is  therefore  seldom  used 
without  modification.  Francis  adopted  this  formula  without 
correcting  for  non-uniform  velocity  in  a  cross-section  of  the  channel 
of  approach,  which  gives  a  a  value  of  unity,  but  this  was  before 
experiments  on  the  effect  of  velocity  of  approach  were  available. 
A  common  method  of  simplification  is  to  simply  drop  the  last 
term  of  formula  (27)  using  as  the  base  formula 

)3A  .......     (29) 


The  value  of  the  term  (ah)**  which  is  dropped  is  represented  by 
the  area  MOS  (Fig.  77).  The  amount  which  the  discharge  is 
affected  by  the  term  ah  as  retained  is  represented  by  the  area 
OSNP.  These  areas  are  purely  illustrative  as  actual  areas  are 
dependent  upon  values  of  H  and  V.  By  substituting  numerical 
values,  however,  it  may  be  shown  that  within  the  range  of  con- 
ditions occurring  in  practice,  the  simplified  formula  is  nearly 
equivalent  to  the  original  expression.  It  should  also  be  noted 
that  a  large  portion  of  the  error  that  would  otherwise  be  introduced 
by  dropping  the  term  (ah)%  may  be  corrected  in  the  selection  of 
coefficients.  The  present  understanding  of  weir  hydraulics  and 
the  experimental  data  available  for  the  determination  of  empi- 
rical coefficients  are  not  sufficient  to  justify  too  close  an  adherence 
to  fundamental  formulas. 

Equation  (29)  is  not  in  a  form  convenient  to  use  since  h  depends 
upon  V  and  therefore  upon  Q  for  its  value.  When  Q  is  unknown  a 
formula  of  this  form  must  be  solved  by  first  determining  the 
approximate  value  of  Q,  neglecting  velocity  of  approach  (formula 
26).  From  this  value  of  Q  an  approximate  value  of  h  may  be 
obtained,  which  substituted  in  the  formula  involving  velocity  of 
approach  correction  gives  a  value  of  Q  which  is  usually  close 
enough  for  the  purpose.  If  a  closer  result  is  desired  the  compu- 
tations may  be  repeated  using  this  new  value  of  Q  for  determin- 
ing h.  This  formula  may  be  modified  by  mathematical  transfor- 
mation so  that  terms  depending  upon  Q  for  their  value  do  not 
occur  on  the  right-hand  side  of  the  equation. 

75.  Algebraic  Transformation  of  Formula.  —  The  fundamental 
formula  as  derived  in  Art.  72  is 


114  FLOW  OF   WATER  OVER  WEIRS 

Expanding  the  left-hand  term  within  the  brackets  by  the  binomial 
theorem  gives  a  diminishing  series,  and  dropping  the  terms  of 
higher  powers  the  equation  becomes 


An  expression  approximately  equivalent  to  the  above  is  obtained 
by  dropping  all  of  the  terms  within  the  brackets  excepting  the  first 
two,  which  gives 


(31) 

As  explained  in  the  preceding  article,  the  value  of  equation  (30)  is 

changed  but  little  by  dropping  the  term  |  -~  ]   and  since  the  sum 

\HI 
of  the  terms  of  the  expanded  series  which  are  dropped  is  of  oppo- 

iah\  K 
site  sign  and  less  than  (  -jj  j    ,  formula  (31),  is  a  closer  approxima- 

tion to  the  fundamental  formula  (27)  than  formula  (29)  . 

It  is  now  desired  to  eliminate  h,  which  depends  upon  Q  for  its 
value.     By  definition  of  velocity  of  approach, 


V  =     =  (approximately),       .     .     .     (32) 

A.  A. 

where  A  is  the  cross-sectional  area  of  the  stream  in  the  channel  of 
approach.  The  value  of  Q  as  substituted  is  an  approximation 
since  it  does  not  include  the  velocity  of  approach  correction. 
It  is  to  be  applied,  however,  to  a  term  which  is  itself  a  small 
correction,  making  the  error  introduced  by  this  approximation 
relatively  unimportant.  Using  this  value  of  V 

V2    C2L2H* 
h=2J-W  .........     (33) 

Substituting  this  value  of  h  in  formula  (31)  and  Deducing, 

.....    (34) 


Replacing  -^  —  by  a  single  coefficient,  Ci,  the  formula  becomes 


(35) 


WEIR  EXPERIMENTS  115 

This  is  a  convenient  base  formula  for  discharge  over  weirs.  Some 
weir  formulas  are  expressed  in  this  form  and  most  of  the  others 
may  be  readily  reduced  to  it.  This  form  of  formula  provides  a 
direct  solution  for  Q  while  other  forms  require  a  trial  and  error 
method  of  correcting  for  velocity  of  approach.  The  values  of 
C  and  Ci  must  be  determined  from  experiments  and  at  this  point 
rational  reasoning  must  give  place  to  empirical  science. 

76.  Weir  Experiments.  —  Working  formulas  are  obtained  by 
determining  experimentally  the  values  of  coefficients  to  be  applied 
to  derived  formulas  as,  for  instance,  C  and  C\  in  (35).     Many 
experiments,  on  sharp-crested  weirs,  covering  a  wide  range  of 
conditions  of  flow,  have  been  performed  during  the  past  century. 
The  most  important  of  these  are  the  experiments  by  Francis  in 
1852,  those  by  Fteley  and  Stearns  in  1877  and  the  Bazin  experi- 
ments in  1886.1     There  are  some  inconsistencies  in  the  results  of 
the  various  experiments,  but  in  general  they  substantiate  the  cor- 
rectness of  the  reasoning  in  the  preceding  pages  and  the  base 
formula  derived  thereby. 

77.  Formulas  for  Sharp-crested  Weirs.  —  A  large  number  of 
formulas  for  sharp-crested  weirs  have  been  published,  but  only 
those  best  known  or  those  appearing  to  possess  the  greatest  merit 
will  be  given. 

The  Francis  Formula  which  is  obtained  by  putting  C  =  3.33 
and  a  =  1  in  formula  (27)  is  as  follows  : 

-hH]    ......     (36) 


Substituting  C  =  3.33  and  a=  1  in  (34)  there  is  obtained  the  follow- 
ing formula  which  gives  results  very  nearly  the  same  as 
formula  (36). 

.     .     .     (37) 

This  may  be  considered  as  another  form  of  the  Francis  formula, 
more  convenient  than  the  original,  since  it  affords  a  direct  solution 

1  J.  B.  FRANCIS:  Lowell  Hydraulic  Experiments  (4th  edition,  1883). 
Also  Trans.  Amer.  Soc.  Civ.  Eng.,  vol.  13,  p.  303..  FTELEY  and  STEARNS: 
Flow  of  Water  over  Weirs.  Trans.  Amer.  Soc.  Civ.  Eng.,  vol.  12  (1883). 
H.  BA/IN:  Annales  des  Fonts  et  Chaussees,  October,  1888.  Translation  by 
MARICHAL  and  TRAUTWINE:  Proc.  Eng.  Club,  Phila.,  January,  1890.  Also 
Annales  des  Fonts  et  Chaussees  for  1894,  ler  Trimestre. 


116  FLOW  OF  WATER  OVER  WEIRS 

for  Q,  whereas  the  form  (36)  requires  the  trial  method  of  solution. 
The  second  term  within  the  brackets  is  the  velocity  of  approach 
correction.  When  the  velocity  of  approach  is  very  small  the  term 


—  - 


0.26    —  j-J   may  be  neglected  and  the  Francis  formula  reduces  to 

(38) 


If  there  are  end  contractions  the  measured  length  of  weir  should 
be  corrected  by  formula  (28)  . 

The  Fteley  and  Stearns  Formula,,  based  upon  a  study  of  their 
own  experiments  and  the  experiments  of  Francis,  is 

Q  =  3.31L(#+«/03/2+0.007L,     ....     (39) 

in  which  a  =  1.5  for  suppressed  weirs  and  2.05  for  weirs  with  end 
contractions.  Without  velocity  of  approach  correction  h  =  Q. 
Formula  (28)  is  to  be  used  to  correct  for  end  contractions 

The  Bazin  Formula.  —  The  experiments  on  suppressed  weirs  by 
Bazin  covered  a  wide  range  of  conditions.  As  a  result  of  his 
investigation  Bazin  devised  a  formula  applicable  to  suppressed 
weirs.  As  originally  published,  the  Bazin  formula  is  expressed  in 
metric  units.  His  base  formula  is  of  the  form  of  (35).  The 
value  of  coefficients  which  he  derived  (see  Art.  72,  page  111), 
expressed  in  English  units,  may  be  written 

0  =  3.248+4^ 
ti 

and 

Ci  =  0.55. 

Substituting  these  values  in  (35)  the  formula  becomes 

248+^)  [i+o.55(£p)2].    .     .     (40) 

It  was  not  intended  by  Bazin  that  this  formula  should  apply 
to  weirs  with  end  contractions,  though  in  the  form  given  above 
it  can  be  so  used,  after  correcting  the  measured  length,  //,  by 
formula  (28).  Without  veloc'ty  of  approach  correction  the  term 
within  the  brackets  becomes  unity. 

For  suppressed  weirs  in  rectangular  channels  where  L  equals 
the  width  of  the  channel  as  well  as  the  length  of  the  weir  and  d 
equals  the  depth  of  water  in  the  channel  of  approach,  the  area  of 


FORMULAS   FOR  SHARP-CRESTED   WEIRS 


117 


water  section  in  the  channel  of  approach  equals  Ld  and  formula 
(40)  becomes 

(07Q\  /  H2\ 

3.  248+^)^1+0.  55^-J.       .     .     (41) 

In  this  form  the  Bazin  formula  applies  more  conveniently  to  sup- 
pressed weirs. 

The  King  Formula  l  is  based  upon  a  study  of  the  experiments  of 
Francis,  Fteley  and  Stearns,  and  Bazin,  from  which  (see  Art.  72, 
page  111)  values  of 

3.34 


~~ 


and 


were  obtained.     Substituting  these  values  the  formula  becomes 


.        .     .     .     (42) 


This  formula  applies  to  weirs  with  and  without  end  contractions. 
If  there  are  end  contractions  the  measured  length  of  weir  is  to  be 
corrected  by  formula  (28).  In  rectangular  channels  where  the 
weir  length  equals  the  width  of  channel,  and  d  equals  the  depth 
.of  water  in  the  channel  of  approach,  formula  (42)  reduces  to 


(43) 


Without  velocity  of  approach,     =="-=^te^=-=^=^s^. 


theses   equals  unity 
formula  becomes 

and   the 
•   .     (44) 

^^^^%%^^^^ 

\ 

Falls    (Fig.    79)    may    be  FIG.  79.  —  Fall. 

considered   as  weirs  having  a 
height  of  zero.     In  this  case  the  head  equals  the  depth  of  water, 


(1918). 


.  W.  KING:    Handbook  of  Hydraulics,  McGraw-Hill  Book  Co.,  p.  71 


118  FLOW  OF  WATER  OVER   WEIRS 

and  if  the  channel  has  vertical  sides,  H/d,  formulas  (41)  and  (43), 
equals  unity.     These  formulas  then  become,  respectively, 


(45) 

and 

(46) 


Both  of  these  formulas  lack   experimental  verification. 

78.  Discussion  of  Weir  Formulas.  —  As  is  the  case  with  all 
empirical  formulas,  weir  formulas  are  no  more  accurate  than  the 
data  upon  which  they  are  based.     The  three  sets  of  experiments 
(see  Art.  76)  on  which  the  above  formulas  are  based  give  results 
which  are  somewhat  conflicting  so  that  no  formula  can  agree  with 
them  all.     The  formula  of  Fteley  and  Stearns  being  based  largely 
upon  the  results  of  their  own  experiments  gives  discharges  some- 
what less  than  the  Bazin  and  King  formulas  which  agree  more 
closely  with  the  Bazin  experiments. 

The  Francis  formula  is  based  entirely  upon  the  Francis  experi- 
ments, which  do  not  cover  a  wide  range  of  conditions  nor  include 
any  measurements  for  a  determination  of  the  effects  of  velocity 
of  approach.  As  a  result  of  this  the  Francis  formula  gives  results 
considerably  in  error  for  high  velocities  of  approach.  It  has  also 
been  found  that  the  Francis  formula  gives  too  small  discharges 
for  low  heads.  For  these  reasons  the  Francis  formula  should  not 
be  considered  as  generally  applicable  to  all  conditions. 

In  general,  formulas  of  the  form  of  (35)  which  do  not  have 
terms  dependent  upon  Q  on  the  right-hand  side  of  the  equation 
are  much  more  convenient  to  use  than  those  of  the  form  of  (27) 
or  (29)  .  There  is  nothing  sacrificed  in  accuracy  by  using  formulas 
of  the  former  type. 

79.  Submerged  Weirs.  —  If  the  elevation  of  the  water  surface 
in  the  channel  below  a  weir  is  higher  than  the  crest  of  the  weir 
the  weir  is  said  to  be  submerged  or  drowned.     The  water  flowing 
away  from  the  weir  is  sometimes  called  the  tail  water.     The  chan- 
nel below  the  weir  is  called  the  channel  of  retreat  and  the  velocity 
in  this  channel  is  the  velocity  of  retreat.     The  depth  of  submergence 
is  the  difference  in  elevation  between  the  tail-water  surface  and 
the  crest  of  the  weir.     Other  terms  used  correspond  to  those  for 
weirs  with  free  overfall. 


SUBMERGED  WEIRS 


119 


Fig.  80  represents  a  submerged  rectangular  weir.  The  head 
is  H  and  the  depth  of  submergence  D.  The  difference  in  elevation 
of  water  surfaces  is  Z  =  H  —  D.  The  length  of  weir  is  L. 

In  Art.  65  a  submerged  weir  is  shown  to  be  a  special  case  of  a 
partially  submerged  orifice.  The  discharge  may  be  considered  as 
the  combined  discharge  of  a  weir  whose  crest  is  at  the  elevation  of 
the  tail  water  and  a  submerged  orifice,  each  discharging  under  a 
head  Z.  Neglecting  velocity  of  approach,  the  combined  discharge  is 


Q  =  %C'V2gLZ/~+C"V2gLDVZ.    .     .     .     (47) 
Writing  C,  for  fC'Vfy  and  C2  for  C"V2g  the  formula  becomes 

(48) 


FIG.  80. — Submerged  weir. 

From  experiments  by  Fteley  and  Stearns  and  by  Francis  are 
obtained  values  of  coefficients  which  substituted  in  formula  (48) 
give  the  following  formula  for  submerged  weirs: 


(49) 


This  formula  does  not  provide  any  method  of  correcting  for 
velocity  of  approach  nor  of  making  other  corrections  explained 
below.  Results  obtained  by  formulas  of  this  type  must  be  con- 
sidered very  approximate  excepting  for  weirs  that  nearly  duplicate 
the  conditions  of  the  original  experiments. 

Some  investigators  have  considered  C'  and  C" ',  formula  (47), 
to  be  of  the  same  value — that  is,  they  have  considered  the  crest 
contraction  to  equal  the  surface  contraction.  If  C'  and  C"  are 
made  equal,  equation  (47)  may  be  reduced  to  the  form 


K)- 


(50) 


120  FLOW  OF  WATER  OVER  WEIRS 

This  is  the  formula  adopted  by  Fteley  and  Stearns.  Accompany- 
ing the  formula  they  give  a  table  of  values  of  C  varying  from 
3.372  to  3.089  for  different  values  of  D/H.  Since  formula  (50) 
requires  an  accompanying  table  of  coefficients,  it  is  not  as  con- 
venient as  formula  (49)  and  possesses  no  advantages  from  con- 
siderations of  accuracy. 

80.  Further  Discussion  of  Submerged  Weirs. — The  discharge 
over  submerged  weirs  is  affected  by  velocity  of  approach  in  a 
manner  very  similar  to  the  discharge  over  weirs  with  free  overfall, 
but  as  the  formula  is  more  complicated,  the  application  of  a 
velocity  of  approach  correction  is  more  difficult.  If  in  deriving 
formula  (45)  (page  95)  the  head  is  increased  by  an  amount  ah 
to  correct  for  velocity  of  approach  and  the  distance  from  the  top 
of  the  orifice  to  the  water  surface  is  made  zero  the  following  formula 
is  obtained : 

Q  =  ^CfLV^[(Z+ah)3/2-(ah)3/~]+Cf/LV2^(Z+ah)l/-(H-Z).     (51) 

This  formula  is  complicated  and  is  not  reducible  to  a  form  permit- 
ting of  simple  application  to  submerged  weir  problems.  With  the 
limited  experimental  data  available  it  is  not  possible  to  obtain 
values  of  the  coefficients  C",  C"  and'a  with  any  degree  of  accuracy. 
There  are  required,  moreover,  other  corrections,  largely  empirical 
in  character  which  if  applied  to  the  above  formula  will  still  further 
complicate  it. 

The  discharge  over  a  submerged  weir  is  greatly  affected  by 
conditions  in  the  channel  below  the  weir.  Water  flows  over  the 
crest  of  the  weir  at  a  velocity  which  is  usually  higher  than  the 
normal  velocity  of  the  tail  water  and  a  portion  of  this  velocity  is 
retained  temporarily  after  leaving  the  weir.  Where  the  slope  of 
the  channel  is  not  sufficient  to  maintain  this  high  velocity  there  is 
a  piling-up  effect  just  below  the  faster-moving  water.  This  condi- 
tion is  illustrated  in  Fig.  80.  The  water  has  a  higher  velocity 
at  a  and  a  lower  velocity  at  b  than  the  normal  velocity  in  the 
channel.  This  condition  produces  what  is  known  as  the  standing 
wave,  a  being  the  trough  and  b  the  crest  of  the  standing  wave. 
Below  the  main  wave  a  series  of  smaller  waves  form,  which  gradu- 
ally reduce  in  size  and  finally  disappear. 

The  factors  affecting  the  height  of  the  standing  wave  are  not 
well  determined,  but  from  a  purely  empirical  investigation  it 


TRIANGULAR  WEIRS  121 

appears  to  increase  directly  as  the  square  noot  of  H,  D  and  Z 
(Fig.  80)  and  inversely  as  the  square  root  of  di.  The  depth  of 
submergence,  D,  is  one  of  the  terms  entering  into  submerged  weir 
formulas  and  its  value  should  be  accurately  determined.  Theo- 
retically, it  is^the  depth,  DI,  in  the  trough  of  the  standing  wave, 
that  is  to  be  used  in  formula  (47)  ,  but  usually  it  is  the  depth  of 
submergence  at  D,  below  all  turbulence  caused  by  water  flowing 
over  the  weir  that  is  most  easily  measured  and  most  convenient  to 
use  in  submerged  weir  problems. 

The  only  experiments  on  submerged  weirs  that  furnish  any  data 
relative  to  the  effect  of  velocity  of  approach  and  channel  conditions 
below  the  weir  are  the  experiments  by  Bazin.  Two  formulas  for 
submerged  rectangular  sharp-crested  weirs  without  end  contrac- 
tions which  accord  very  well  with  the  results  of  the  Bazin  experi- 
ments are  given  below.  These  formulas  are  largely  empirical  in 
character  and  a  discussion  of  their  derivation  is  not  given.  The 
symbols  used  are  indicated  in  Fig.  80.  The  submerged  weir 
formula  by  Bazin  1  may  be  written 


K 

The  submerged  weir  formula  by  King  2  is 


.     (52) 


(53) 


Each  of  the  above  formulas  will  require  further  experimental 
verification  before  it  can  be  considered  applicable  to  all  conditions. 

81.  Triangular  Weirs.— 
Fig.  81  represents  a  trian- 
gular notch  or  weir  over 
which  water  is  flowing.  The 
measured  head  is  H  and 

the    distance    betweeii    the 

•  j          r    ,T_  •   •  ~,     ,1  FIG.  81.  —  Triangular  weir. 

sides    of   the    weir    at   the 

water  surface  is  /.     The  sides  make  equal  angles  with  the  vertical. 
The  area  of  an  elementary  horizontal  strip  dy  in  thickness  is 

1  Annales  des  Fonts  et  Chaussees  for  1898,  ler  Trimestre,  p.  235. 

2  Handbook  of  Hydraulics,  p.  82.  ^ 


122  FLOW  OF  WATER  OVER  WEIRS 

I'dy.  Neglecting  velocity  of  approach,  the  theoretical  velocity 
through  this  strip  for  a  head  y  is  V2gy  and  the  theoretical 
discharge  is 


from  similar  triangles 


H 

Combining  the  two  equations 


integrating  between  the  limits  H  and  0  and  reducing, 

(54) 


The  slope  which  the  sides  of  the  weir  make  with  the  vertical  may 
be  represented  by  z,  then 


=  z    or    l  =  2zH.  (55) 

n 

Substituting  this  value  of  I  in  (54)  the  theoretical  formula  for 
discharge,  expressed  in  terms  of  head  and  slope  of  sides,  is, 


(56) 

Applying  a  coefficient  of  discharge  and  combining  it  with  the 
constant  terms  the  same  as  for  rectangular  weirs 

Q  =  CzH*,    .........     (57) 

in  which  the  value  of  C  must  be  determined  experimentally. 

If  the  angle  between  the  sides  is  a  right  angle,  z  equals  unity. 
Most  of  the  available  experimental  data  are  for  right-angled 
notches.  Triangular  weirs  having  other  angles  are  seldom  used. 
Experiments  indicate  quite  clearly  that  C  is  not  a  constant,  its 
value  decreasing  with  increasing  heads. 

The  following  are  values  of  C  as  obtained  from  various  sets  of 
experiments  l  together  with  corresponding  formulas  for  sharp- 

1PROF.  JAMES  THOMPSON:  Experiments  on  Triangular  Weirs.  British 
Association  Reports,  1861.  JAMES  BARE:  Flow  of  Water  over  Triangular 
Notches.  Engineering,  April  8  and  15,  1910.  H.  W.  KING:  Handbook  of 
Hydraulics,  p.  86,  1918. 


TRAPEZOIDAL  WEIRS  123 

edged,  right-angled,  triangular  weirs.     From  Thompson's  experi- 
ments 

C(mean)=2.54        Q  =  2.54ff**.'.     .     .     .     (58) 

From  Barr's  experiments; 

9  48 

-«.       .    .    .    (59) 


From  experiments  at  the  University  of  Michigan; 

'.      .     .     .     (60) 


Other  experiments  on  triangular  weirs  give  results  varying  some- 
what from  those  listed  above. 

One  interesting  fact  brought  out  by  Barr's  experiments  is  that 
the  discharge  over  a  sharp-edged,  metallic,  triangular  weir  may  be 
2  per  cent  greater  when  the  inner  face  of  the  metal  is  rough  than 
when  it  is  smooth.  The  rougher  face,  by  retarding  the  movement 
of  water  parallel  to  it,  reduces  the  velocities  which  have  the  greatest 
influence  on  contraction,  thus  reducing  the  contraction  and  so 
increasing  the  discharge. 

The  effect  of  velocity  of  approch  on  triangular  weirs  is  similar 
in  character  to  the  effect  on  rectangular  weirs.  No  data  for 
determining  coefficients  are,  however,  available.  From  the  nature 
of  the  triangular  weir  the  cross-sectional  area  of  the  nappe  is  usually 
very  much  smaller  than  that  of  the  channel  of  approach.  The 
velocity  of  approach  is  therefore  small  and  the  error  introduced  by 
neglecting  it  is  usually  inappreciable.  This  has  been  confirmed 
experimentally  by  Barr. 

The  triangular  weir  affords  an  excellent  method  of  measuring 
small  discharges.  Formula  (59)  probably  applies  more  accurately 
to  sharp-edged  notches  cut  in  very  smooth  metal  and  (60)  to 
sharp-edged  notches  cut  in  rougher  metal,  such  as  ordinary  com- 
mercial steel  plate. 

For  angles  slightly  greater  or  less  than  90°  it  is  probable  that 
the  values  of  C  listed  above,  if  substituted  in  formula  (57),  will 
give  quite  accurate  results. 

82.  Trapezoidal  Weirs.  —  Fig.  82  represents  a  trapezoidal 
weir  having  a  horizontal  crest  of  length  L.  The  sides  are  equally 
inclined,  making  angles  a/H  =  z  with  the  vertical. 


124  FLOW  OF  WATER  OVER  WEIRS 

By  writing  the  equation 


HH 

-"- 

x 

— 

~~$*~*  JH  

—  y" 

\ 

^dy  I 

(\ 

FIG.  82. — Trapezoidal  weir. 


and  expressing  I'  in  terms  of  y 
and  known  quantities  in  a 
manner  similar  to  that  used 
for  .triangular  weirs,  and  in- 
tegrating and  reducing,  the 
following  formula  for  the  the- 
oretical discharge  over  trape- 

zoidal   weirs    without    velocity  of    approach  correction  can  be 

obtained  : 

Q,-!v^LHM+&Vty*H«  .....    (61) 

The  same  formula  is  obtained  directly  by  the  addition  of  the 
theoretical  discharges  over  rectangular  and  triangular  weirs. 
With  coefficients  included  the  formula  for  discharge  may  be  written 

Q  =  CiLH3A+C2zH5/2  ......     (62) 

There  are  no  experimental  data  for  the  determination  of  C\  and  €2 
and  for  this  reason  the  trapezoidal  weir  has  little  practical  value. 

83.  The  Cippoletti  Weir.  —  A  trapezoidal  weir,  having  a  value 
of  z  =  a/H  (Fig.  82)  of  J,  is  called  a  Cippoletti  1  weir.  This  slope 
of  the  sides  is  approximately  that  required  to  secure  a  discharge 
through  the  triangular  portion  of  the  weir  opening  that  equals  the 
decrease  in  discharge  resulting  from  end  contractions.  The 
advantage  of  the  Cippoletti  weir  is  that  it  does  not  require  a 
correction  for  end  contractions.  The  method  employed  by 
Cippoletti  in  arriving  at  his  value  of  z  is  as  follows  : 

The  discharge  through  the  triangular  portion  of  the  weir, 
C"  being  the  coefficient  of  discharge,  is 


The  decrease  in  discharge  resulting  from  end  contractions,  C" 
being  the  coefficient  of  discharge,  according  to  Francis  is 


Equating  the  right-hand  members  of  these  equations,  assuming 
C'  to  equal  C",  and  reducing,  there  results 

«  =  i  .........     (63) 

1  C.  CIPPOLETTI:  Canal  Villoresi  (1887).     Description  of  trapezoidal  weir. 


WEIRS   NOT  SHARP  CRESTED 


125 


The  formula  given  by  Cippoletti  for  determining  the  discharge 
over  Cippoletti  weirs  is 

......     (64) 


to  be  corrected  for  velocity  of  approach  by  the  Francis  method. 
Later  experiments  indicate  that  this  formula  gives  too  great  dis- 
charges for  the  higher  heads  when  the  velocity  of  approach  is  low. 
The  Cippoletti  weir  is  used  quite  extensively  in  western  United 
States  for  measuring  irrigation  water  where  precision  in  measure- 
ment is  not  essential. 

84.  Weirs  Not  Sharp  Crested.  —  Weirs  having  cross-sections 
such  that  they  partially  or  completely  suppress  contractions  at  the 


^"^          * 


FIG.  83. — Weir  with  rectangular 
cross-section,  Nappe  springing 
clear. 


FIG.  84. — Weir  with  rectangular  cross- 
section,  Nappe  adhering. 


FIG.  85. — Weir  with  rounded  crest 


FIG.  86. — Weir  with  Ogee  cross-section. 

crest  are  used  frequently  in  hydraulic  structures.  Spillway  sec- 
tions of  dams  are  examples  of  this  type 
of  weirs.  Such  weirs  also  may  be  used 
as  a  means  of  measuring  water  if 
coefficients  for  the  particular  shape  of 
weir  are  available. 

Figs.  83   to   88   illustrate  various 

cross-sections  of  weirs.     Figs.  83  and  FlG  87  J_Weir  with  triangul 
84    have    rectangular    sections    with  cross-section, 

sharp     upstream     corners.       If      the 
breadth  of   weir,   b  (Fig.  83),  is   about   \E   or   less   the  nappe 


ar 


126 


FLOW  OF  WATER  OVER   WEIRS 


%/   '  •* 

/      A        \l 


-3-0- 


.._6T0 — 


6  to 


Directions  of  flow,  left  to  right 


FIG,  88. — Various  sections  of  weirs  and  dams. 


WEIRS   NOT   SHARP   CRESTED 


127 


HORTON'S  VALUES  OF  WEIR  COEFFICIENT,  C.     (See  Fig.  88) 


Cross-section 

HEAD  IN  FEET,  H 

0.5 

1.0 

1.5 

2.0 

2.5 

3.0 

3.5 

4.0 

4.5 

5.0 

A 

3.46 

3.45 

3.42 

3.35 

3.32 

3.33 

3.37 

3.41 

3.44 

B 

3.43 

3.39 

3.38 

3.38 

3.39 

3.40 

C 

3.26 

3.28 

3.32 

3.38 

3.47 

3.53 

3.59 

3.63 

3.66 

D 

3.29 

3.29 

3.32 

3.36 

3.40 

3.43 

3.48 

3.53 

3.62 

3.72 

E 

3.27 

3.38 

3.46 

3.51 

3.55 

3.58 

3.61 

3.67 

3.74 

3.83 

F 

3.15 

3.45 

3.63 

3.75 

3.82 

3.87 

3.88 

3.88 

G 

3.18 

3.30 

3.38 

3.42 

3.46 

3.49 

3.52 

3.53 

H 

3.28 

3.50 

3.54 

3.52 

3.36 

3.31 

3.30 

3.30 

I 

3.18 

3.27 

3.43 

3.52 

3.59 

3.64 

3.68 

3.70 

J 



3.44 

3.35 

3.30 

3.32 

3.37 

3.38 

3.39 

3.39 

K 

3.12 

3.20 

3.22 

3.22 

3.22 

3.22 

3.22 

3.22 

3.22 

L 

3.12 

3.14 

3.10 

3.14 

3.20 

3.26 

3.21 

3.36 

M 

3.80 

N 

3.10 

3.10 

3.33 

0 

3.53 

3.54 

3.54 

3.49 

3.35 

3.27 

3.25 

3.25 

P 

2.81 

2.81 

2.81 

2.81 

2.81 

2.81 

2.81 

2.81 

2.81 

Q 

3.49 

3.50 

3.52 

R 



3.72 

3.82 

3.85 

3.82 

3.76 

3.68 

3.68 

3.73 

3.82 

S 

3.58 

3.56 

3.57 

3.58 

3.60 

3.62 

3.65 

3.68 

T 

2.70 

2.64 

2.64 

2.70 

2.80 

2.89 

U 

2.72 

2.64 

2.64 

2.64 

2.64 

2.64 

V 

2.72 

2.63 

2.63 

2.63 

2.63 

2.63 

2.63 

2.63 

2.63 

2.63 

I        ! 

will  spring  clear  of  the  downstream  edge  and  there  will  be  com- 
plete crest  contraction.  In  this  case  the  discharge  will  be  given 
by  the  formula  for  sharp-crested  weirs.  If  the  breadth,  b,  is 
such  that  the  nappe  does  not  spring  clear,  as  is  indicated  in 
Fig.  84,  the  free  fall  of  the  nappe  is  interfered  with  and  the  dis- 
charge is  less  than  that  of  a  sharp-crested  weir.  In  Fig.  85  the 
upstream  edge  of  the  weir  is  rounded,  which  reduces  crest  con- 
traction and  thereby  increases  the  discharge.  By  proper  design 
the  crest  contraction  may  be  reduced  very  nearly  to  zero. 

The  base  formula  commonly  used  for  weirs  not  sharp  crested 
(see  Art.  72)  is 

Q  =  CLH%, (26) 

in  which  C  is  a  coefficient  varying  with  H  whose  value  must  be 
determined  at  different  heads  for  each  shape  of  crest. 


128  FLOW  OF  WATER  OVER  WEIRS 

After  a  thorough  investigation  Horton  l  has  prepared  tables 
and  curves  of  C,  corresponding  to  different  heads,  for  practically 
all  shapes  of  weir  sections  for  which  experimental  data  are  avail- 
able. In  computing  the  values  of  his  coefficients  Horton  assumed 
the  velocity  of  approach  correction  given  by  the  formula 

......     (65) 


This  formula  is  obtained  from  formula  (27)  by  giving  a  a  value 
of  unity  and  dropping  the  last  term.  The  correction  is  doubtless 
too  small,  but  since  it  was  used  in  reducing  the  experimental  data 
it  should  be  applied  in  weir  problems  where  Horton's  coefficients 
are  used.  By  substituting  a=l  in  equation  (34)  and  reducing 

there  is  obtained 

r  /r.fj\2~\ 

....     (66) 

which  gives  results  practically  equivalent  to  (65)  and  is  more 
convenient  to  use. 

Fig.  88  shows  various  sections  of  weirs  and  dam  crests  for 
which  experimental  data  are  available.  The  table  on  page  127 
gives  Horton's  values  of  C  for  these  shapes. 

The  degree  of  accuracy  which  may  be  expected  from  the  use 
of  weirs  not  sharp  crested  depends  upon  the  experimental  data 
available  for  determining  C.  Inasmuch  as  there  are  innumerable 
shapes  that  may  be  used,  it  is  not  probable  that  experimental 
data  for  any  large  number  of  them  will  be  secured  for  many 
years.  Complete  data  for  any  particular  shape  of  weir  requires 
an  exhaustive  research  similar  to  that  required  for  sharp-crested 
weirs.  The  data  at  present  available  are,  however,  sufficient  to 
assist  in  the  selection  of  approximate  coefficients  for  the  shapes 
of  weirs  commonly  used  in  hydraulic  design. 

Weirs  not  sharp  crested,  having  cross-sections  similar  to  the 
shapes  for  which  experimental  values  of  coefficients  are  available, 
may  be  used  for  the  approximate  measurement  of  discharges. 
There  are  some  cross-sectional  forms  which  might  be  more  satis- 
factory for  the  measurement  of  flowing  water  than  sharp-crested 
weirs  if  as  complete  experimental  data  for  them  were  available. 

1  ROBERT  E.  HORTON:  Weir  Experiments,  Coefficients  and  Formulas. 
Water  Supply  and  Irrigation  Paper,  No.  200,  U.  S.  Geol.  Survey  (1907). 


BROAD-CRESTED   WEIRS  OR  CHUTES 


129 


Existing  dams  frequently  may  be  used  for  estimating  flood 
discharges  of  streams  where  direct  measurements  of  discharge  by 
other  methods  are  impracticable. 

85.  Broad-crested  Weirs  or  Chutes. — A  weir  having  a  broad 
flat  top  such  as  is  illustrated  in  Fig.  89  is  called  a  broad-crested 
weir.  A  broad-crested  weir  is  usually  understood  to  be  a  weir 
having  an  approximately  rectangular  cross-section  with  a  width 
of  top,  b  (Figs.  84  and  89)  great  enough  to  prevent  the  nappe 
from  springing  clear  of  the  top  of  the  weir.  Fig.  89  may  also 
be  considered  to  represent  a  longitudinal  section  of  a  chute, 
that  is,  a  short  channel  discharging  from  a  body  of  comparatively 
still  water.  The  chute  bears  a  relation  to  the  weir  analogous 
to  the  relation  of  the  short  tube  to  the  orifice  (page  85).  A 


FIG.  89.  —  Broad-crested  weir. 

rational  derivation  of  a  formula  for  discharge  over  broad-crested 
weirs  or  chutes  is  given  below. 

There  will  be  a  drop,  h,  Fig.  89,  in  the  water  surface  near 
the  upstream  edge  of  the  crest.  The  velocity  of  water  below 
this  drop  is  that  due  to  the  head,  h,  or 


where  v  is  the  mean  velocity  and  Cv  is  the  coefficient  of  velocity. 
If  the  top  of  the  weir  is  level  or  has  a  very  gentle  slope  the  depth, 
d,  will  remain  very  nearly  constant  from  the  place  where  h  is 
measured  to  the  lower  edge  of  the  crest.  With  a  greater  slope 
of  crest  the  velocity  will  accelerate  and  d  will  gradually  decrease 
toward  the  lower  edge  of  the  crest.  The  discharge  will  not, 
however,  be  materially  affected  by  the  slope  of  the  crest,  pro- 
vided it  is  sufficient  to  maintain  the  velocity  v,  since,  as  is  shown 
below  there  is  a  maximum  discharge  which  can  not  be  exceeded. 


130  FLOW  OF  WATER  OVER  WEIRS 

If  L  is  the  length  of  weir  or  width  of  chute,  the  area  through 
which  water  is  discharging  under  a  head,  h,  is 

a  =  Ld  =  L(H-h'). 

The  mean  velocity  is  that  due  to  the  head,  h,  multiplied  by  the 
coefficient  of  velocity  Cv,  and  the  discharge  is 


The  coefficient  of  contraction  is  unity  and  Cv,  therefore,  is  also 
the  coefficient  of  discharge. 

In  this  equation  Q  =  0  when  h  =  0  and  also  when  h  =  H. 
There  is  therefore  an  intermediate  value  of  h  which  makes  Q  a 
maximum.  This  maximum  value  of  Q  can  be  obtained  by 
differentiating  and  equating  to  zero,  which  gives 


whence, 

h  =  \H  ............     (68) 

Substituting  this  value  of  h  in  equation  (67)  and  reducing  gives 
Q  =  3.087  CVLHK  .......     (69) 


The  coefficient,  Cv,  is  similar  in  character  to  the  coefficient  of 
velocity  for  a  standard  short  tube,  Fig.  60.  Its  value  depends 
upon  the  shape  of  the  upstream  edge  of  the  crest  and  probably 
approaches  a  maximum  value  of  about  0.98  when  this  edge  is 
so  rounded  as  to  prevent  contraction.  Formula  (69)  may  also 
be  written 

.......     (26) 


which  is  the  base  formula  for  weirs  not  sharp  crested. 

From  experiments  on  broad-crested  weirs  it  has  been  found 
that  for  weirs  having  a  breadth  of  10  ft.  or  more,  discharging 
under  a  head  of  1.0  ft.  or  more, 


(70) 


which  corresponds  to  a  coefficient,  Cv,  in  formula  (69)  of  0.85. 
If  there  are  end  contractions  a  separate  correction  must  be 
applied  to  the  length. 


MEASUREMENT  OF  HEAD 


131 


Formula  (69)  is  of  fundamental  importance  in  connection 
with  the  entrance  conditions  for  open  channels.  It  gives  the 
maximum  rate  at  which  water  can  be  drawn  through  an  open 
channel  from  any  body  of  comparatively  still  water.  The  rate 
of  discharge  may  be  less  but  it  can  never  be  more  than  that 
given  by  the  formula  (see  Art..  126). 

86.  Measurement  of  Head. — In  using  weirs  to  measure  the 
rate  of  discharge,  the  head,  length  of  weir  and  cross-sectional 
area  of  the  channel  of  approach  must  be  carefully  measured. 
The  last  two  of  these  usually  need  to  be  measured  but  once  and 
can  then  be  used  in  all  subsequent  determinations  of  Q. 

The  head  is  measured  with  some  form  of  a  gage  which  is  set 
in  a  fixed  position.  The  elevation  of  the  zero  of 
the  gage  with  reference  to  the  crest  of  the  weir 
should  be  accurately  determined.  It  is  preferable 
to  measure  the  head  in  a  well  or  stilling  box  con- 
nected to  the  channel  by  a  small  pipe,  the  end  of 
which  is  flush  with  the  side  of  the  channel.  This 
provides  a  means  for  measuring  the  head  in  still 
water  and  reduces  the  effect  of  waves  which  are 
usually  present  in  the  channel  of  approach.  For 
the  most  precise  work  a  hook  gage  should  be  used. 

The  hook  gage,  Fig.  90,  consists  of  a  graduated 
metallic  rod  with  a  pointed  hook  at  the  bottom 
which  slides  vertically  in  fixed  supports.  By  means 
of  a  vernier  attached  to  one  of  the  supports,  read- 
ings to  thousandths  of  a  foot  may  be  taken.  The 
rod  usually  has  a  range  of  movement  of  about 
2  ft.  The  gage  should  be  rigidly  attached  to  a 
support  at  such  an  elevation  that  the  movement 
of  the  hook  covers  the  range  of  water  surface  eleva- 
tions to  be  read.  To  take  a  reading,  the  point  of 
the  hook  is  lowered  below  the  surface  and  then 
slowly  raised  by  the  screw  at  the  top  of  the  instru- 
ment. Just  before  the  point  of  the  hook  pierces 
the  skin  of  the  water,  a  pimple  is  seen  on  the  FIG.  90. 
surface;  the  hook  is  then  lowered  slightly  until  the  H°°k  gage, 
pimple  is  barely  visible  and  the  vernier  is  read. 

Where  less  precision  is  required,  especially  for  securing  con- 
tinuous records  of  elevation  as  in  ordinary  stream  gaging  work, 


132  FLOW  OF  WATER  OVER  WEIRS 

some  other  form  of  gage  is  desirable.  An  ordinary  staff  gage, — 
that  is,  a  painted  rod  graduated  to  feet  and  decimals  of  a  foot 
so  set  that  the  water  surface  comes  in  contact  with  the  gradua- 
tions— is  quite  satisfactory  in  some  cases. 

There  are  a  great  many  different  types  of  recording  gages 
which  give  continuous  records  of  water-surface  elevation.  These 
gages  either  provide  a  record  by  a  graph,  the  coordinates  of  which 
indicate  the  time  and  stage,  or  by  a  device  that  prints  elevations 
at  stated  intervals  of  time.  The  essential  parts  of  a  recording 
gage  are:  a  float  which  rises  and  falls  with  the  surface  of  the 
water,  a  device  for  transferring  the  vertical  motion  of  the  float 
to  the  record,  a  recording  device,  and  a  clock. 

Another  device  for  determining  head  is  a  plummet  attached 
to  the  end  of  a  steel  tape.  This  is  used  to  measure  the  vertical 
distance  from  a  fixed  point  above  the  channel  of  approach  to  the 
water  surface.  The  reading  of  the  tape  when  the  point  of  the 
plummet  is  at  the  elevation  of  the  crest  of  the  weir  is  first  deter- 
mined accurately  and  the  difference  between  this  reading  and 
the  reading  when  the  point  just  touches  the  water  surface  gives 
the  head  on  the  weir.  This  method  gives  accurate  results,  but 
for  precise  work  it  probably  is  preferable  to  measure  the  head  in 
a  stilling  box  with  a  hook  gage,  so  as  to  conform  to  the  conditions 
of  the  experiments  upon  which  weir  formulas  are  based. 

The  head  always  should  be  measured  far  enough  upstream 
from  the  weir  to  be  well  above  the  effects  of  surface  contraction. 
In  their  experiments,  Francis,  and  Fteley  and  Stearns  measured 
heads  6  ft.  and  Bazin  16.4  ft.  upstream  from  the  weir.  The 
distance  selected  should  preferably  conform  approximately  to 
that  used  in  the  experiments  on  which  the  formula  to  be  used  in 
computing  discharges  is  based,  though  accurate  comparative 
measurements  show  an  almost  imperceptible  difference  -between 
heads  measured  6  ft.  and  those  measured  16.4  ft.  from  the  weir. 

87.  Conditions  for  Accurate  Measurement  over  Sharp-crested 
Weirs. — To  obtain  maximum  accuracy  the  face  of  the  weir 
should  be  vertical  and  the  crest  level.  The  crest  should  be  cut 
from  plate  metal,  trile  to  line  with  a  flat  top  and  sharp  upstream 
corner. 

Suppressed  weirs  having  a  length  equal  to  the  channel  width 
have  a  space  below  the  nappe  which  may  have  no  connection 
with  the  outside  air.  In  passing  over  this  space  the  nappe 


PROBLEMS  133 

carries  with  it  all  or  a  portion  of  the  entrapped  air,  thus  reducing 
the  pressure  underneath  and  causing  the  nappe  to  be  depressed. 
This  is  equivalent  to  reducing  crest  contraction  making  the 
usual  formulas  inapplicable.  The  space  under  the  nappe, 
therefore,  should  be  connected  by  pipes  or  by  other  means  with 
the  outside  air. 

In  general  all  conditions  such  as  dimensions  of  weir  and 
channel  and  ranges  of  head  should  conform  as  nearly  as  prac- 
ticable to  the  conditions  of  the  experiments  which  form  the  basis 
of  the  formula  that  is  to  be  used  in  computing  discharges.  The 
length  of  the  weir  should  be  at  least  three  times  the  measured 
head.  Heads  less  than  0.2  ft.  are  undesirable  since  very  low 
heads  create  a  tendency  for  the  nappe  to  adhere  to  the  weir 
crest  thus  affecting  the  coefficient  of  contraction.  Though  it  has 
not  been  definitely  proved,  it  appears  from  rather  limited  experi- 
mental data  that  weir  formulas  apply  as  accurately  for  heads 
up  to  4  ft.  as  for  lower  heads. 

Weirs  with  end  contractions  should  have  their  ends  at  a 
distance  of  at  least  two  times  the  head  from  the  sides  of  the 
channel  in  order  to  insure  complete  contraction. 

PROBLEMS 

1.  A  sharp-crested  weir  4  ft.  high  extends  across  a  rectangular  channel 
12  ft.  wide.     If  the  measured  head  is  1.22  ft.,  determine  the  discharge,  using 
formulas  (36),  (37),  (40)  and  (42). 

2.  Solve  Problem  1,  changing  the  height  of  weir  to  2  ft.  and  the  measured 
head  to  1.54  ft. 

3.  Solve  Problem  1,  changing  the  height  of  weir  to  2  ft.  and  the  measured 
head  to  0.25  ft. 

4.  What  length  of  weir  should  be  constructed  in  a  stream  100  ft.  wide 
so  that  the  measured  head  will  be  1.50  ft.  when  the  discharge  is  120  cu.  ft. 
per  second? 

6.  A  rectangular  channel  20  ft.  wide  has  a  3-ft.  depth  of  water  flowing 
with  a  mean  velocity  of  2.45  ft.  per  second.  Determine  the  height  of  sharp- 
crested  suppressed  weir  that  will  increase  the  depth  in  the  channel  of  approach 
to  5  ft. 

6.  A  sharp-crested  weir  2.5  ft.  high  is  built  across  a  rectangular  flume 
30  ft.  wide.     The  measured  head  is  1.25  ft.     In  the  flume  is  another  sharp- 
crested  weir  having  a  height  of  3.5  ft.,  the  middle  of  the  weir  being  on  the 
center  line  of  the  flume.     If  the  Measured  head  on  the  latter  weir  is  1.62  ft. 
what  is  the  length  of  crest? 

7.  A  rectangular,  sharp-crested  weir  is  to  be  consructed  in  a    stream  in 
which  the  discharge  varies  from  2  cu.  ft.  per  second  to  50  cu.  ft.  per  second. 


134  FLOW  OF  WATER  OVER  WEIRS 

Determine  a  length  of  weir,  such  that  the  measured  head  will  never  be  less 
than  0.2  ft.  nor  greater  than  one-third  of  the  length  of  weir. 

8.  Determine  the  discharge  over  a  right-angled,  triangular  weir  if  the 
measured  head  is  1.82  ft. 

9.  The  discharge  over  a  right-angled,  triangular  weir  is  7.28  cu.  ft.  per 
second.     What  is  the  measured  head? 

10.  A  channel  is  carrying  10  cu.  ft.  per  second  of  water.     Assuming  that 
an  error  of  0.005  ft.  may  be  made  in  measuring  the  head,  determine  the 
percentage  of  error  resulting  from  the  use  of  a  right-angled,  triangular  weir, 
and  also  from  the  use  of  a  rectangular  weir  10  ft.  long. 

11.  The  measured  discharge  over  a  dam  100  ft.  long  is  520  cu.  ft.  per 
second  when  the  head  is  1.28  ft.     Determine  the  weir  coefficient  for  this 
head. 

12.  If  in  a  certain  channel  the  velocity  varies  uniformly  from  3  ft.  per 
second  at  the  surface  to  1  ft.  per  second  at  the  bottom,  determine  the  corre- 
sponding value  of  a. 

13.  An    overflow    masonry  dam  is  to  be  constructed  across  a  stream. 
The  stream  is  estimated  to  have  a  maximum  flood  discharge  of  30,000  cu.  ft. 
per  second,  when  the  elevation  of  water  surface  at  the  dam  site  is  1132.0. 
Six  sluice  gates  each  8  ft.  high  and  6  ft.  wide  (C  =  0.85)  are  to  be  constructed 
in  the  dam  with  their  sills  at  elevation  1122.5.     The  main  overflow  weir  for 
which  C  =  2.63  will  be  200  ft.  long  with -a  crest  elevation  of  1184.0.     An 
auxiliary  weir  600  ft.  long  with  a  crest  elevation  of  1185.3  will  operate  during 
floods.     For  this  weir  C  =  3.40.     With  all  sluice  gates  open  what  will  be  the 
elevation  of  the  water  surface  upstream  from  the  weir  when  the  discharge 
is  30,000  cu.  ft.  per  second?     Neglect  velocity  of  approach. 

14.  A  submerged  sharp-crested  weir  2.5  ft.  high  extends  clear  across  a 
channel  having  vertical  sides  and  a  width  of  10  ft.     The  depth  of  water  in  the 
channel  of  approach  is  4.0  ft.,  and  35  ft.  downstream  from  the  weir  the  depth 
of  water  is  3.0  ft.     Determine  Q  by  formulas  (52)  and  (53). 

15.  A  channel  20  ft.  wide  with  vertical  sides  is  carrying  400  cu.  ft.  per 
second  of  water  at  a  depth  of  4.0  ft.     How  high  a  sharp-crested  weir  should 
be  constructed  across  the  channel  to  raise  the  elevation  of  the  water  surface 
0.5  ft.? 


CHAPTER  IX 
FLOW  OF  WATER  THROUGH  PIPES 

88.  Description   and   Definitions. — As   the   term   is  used   in 
hydraulics,  a  pipe  may  be  denned  as  a  conduit  which  carries 
water  under  pressure.     More   commonly  pipes  are   of  circular 
cross-section,  and  hydraulic  formulas  for  the  flow  of  water  through 
pipes  are  usually  expressed  in  a  form  particularly  adaptable  to 
circular  pipes,  but  the  same  general  laws  apply  regardless  of  the 
cross-sectional  shape  of  the  pipe. 

Pipes  which  do  not  flow  full  or  which  flow  full  without  exerting 
pressure  against  the  top  of  the  pipe  are  classed  as  open  channels 
and  are  treated  in  a  separate  chapter  (Chapter  X).  A  city 
water  main  carries  water  under  pressure  and  is  therefore  an 
example  of  a  pipe  while  a  sewer  which  normally  does  not  carry 
water  under  pressure  is  classed  as  an  open  channel. 

Since  friction  losses  in  pipes  are  independent  of  pressure 
(Art.  96)  the  same  laws  apply  to  the  flow  of  water  both  in  pipes 
and  open  channels,  and  the  formulas  for  each  take  the  same 
general  form.  Some  formulas  are  designed  to  be  used  either  for 
pipes  or  open  channels,  but  the  more  common  practice  is  to  use 
different  formulas  for  the  two  classes  of  conduits. 

89.  Wetted  Perimeter  and  Hydraulic  Radius.— The  wetted 
perimeter  of  any  conduit  is  the  line  of  intersection  of  its  wetted 
surface  with  a  cross-sectional  plane.     Thus  for  a  pipe  flowing 
full,  d  being  the  diameter,  the  wetted  perimeter  is  equal  to  the 
circumference  or  ird,  if  flowing  half  full  it  is  \ird. 

The  hydraulic  radius  of  a  conduit  is  the  area  of  a  cross-section 
of  the  stream  divided  by  the  wetted  perimeter  of  that  section. 
For  a  circular  pipe  flowing  either  full  or  half  full  the  hydraulic 
radius,  r,  is  evidently  d/4  or  R/2,  R  being  the  radius  of  the  pipe. 

The  terms  wetted  perimeter  and  hydraulic  radius  are  used 
more  generally  in  connection  with  open  channels  than  with 
pipes,  but  they  are  sometimes  used  in  pipe  formulas.  Their 
application  to  open  channels  is  discussed  in  Art.  109. 

135 


136  FLOW  OF  WATER  THROUGH  PIPES 

90.  Critical  Velocities  in  Pipes. — Under  the  conditions  ordi- 
narily encountered  in  hydraulic  practice,  water  flows  through 
pipes  with  a  turbulent  motion,  that  is,  the  water  particles  have 
a  transverse  as  well  as  a  longitudinal  motion,  and  any  particle 
near  the  center  of  the  pipe  at  one  time  may  be  near  its  sur- 
face an  instant  later,  occupying  successively  various  trans- 
verse positions,  while  it  is  at  the  same  time  being  propelled 
forward. 

Though  at  any  instant  the  water  particles  in  a  pipe  where 
turbulent  motion  exists  move  forward  with  different  velocities 
(Art.  91),  the  average  longitudinal  velocity  of  every  particle  is 
approximately  the  same.  This  may  be  shown  by  suddenly  in- 
jecting a  colored  liquid  into  a  pipe  and  observing  the  coloring 
matter  where  it  discharges  from  the  pipe.  It  will  be  observed 
that  the  coloring  matter  remains  in  a  short  prism  even  after  the 
water  has  traveled  a  distance  of  1000  diameters  or  more  and  that 
the  water  on  either  side  of  this  prism  is  comparatively  clear. 
This  principle  is  made  use  of  in  measuring  the  velocity  of  flow 
through  pipes. 

At  comparatively  low  velocities,  water  may  be  made  to  flow 
through  small  pipes  without  turbulence,  that  is  with  stream 
line  motion.  Under  these  conditions  the  water  particles  all 
flow  in  paths  parallel  to  the  axis  of  the  pipe.  The  particles  near 
the  axis  then  flow  with  a  higher  velocity  than  the  other  particles, 
the  velocities  gradually  becoming  less  as  the  distance  from  the 
center  of  the  pipe  increases,  the  lowest  velocity  being  near  the 
surface  of  the  pipe.  This  retardation  of  velocities  is  caused  by 
the  viscosity  of  the  water  and  friction  between  the  moving  water 
and  the  pipe. 

The  flow  of  water  in  small  glass  tubes  has  been  studied  experi- 
mentally by  Reynolds l  in  the  following  manner.  Water  was 
drawn  through  the  tubes  from  a  glass  tank  in  which  the  water 
had  been  allowed  to  come  to  rest,  arrangements  being  made  to 
introduce  threads  of  colored  water  into  the  entrance  of  the  tubes. 
Reynolds  found,  when  the  velocities  were  sufficiently  low,  that 
the  streak  of  color  extended  as  a  beautiful  straight  line  through 
the  tube.  As  the  velocity  of  the.  water  was  increased  by  small 
stages,  a  velocity  was  finally  reached  where  the  color  suddenly 

REYNOLDS:   Phil.  Trans.  Royal  Society,  1882  and  1895. 


CRITICAL  VELOCITIES   IN   PIPES 


137 


mixed  with  the  surrounding  water.  The  velocity  at  which 
mixing  began  was  evidently  the  velocity  at  which  stream-line 
motion  ended  and  turbulent  motion  began.  It  has  been  termed 
the  higher  critical  velocity. 

Reynolds  also  found  that  below  a  certain  limiting  velocity, 
when  the  water  was  disturbed  it  soon  resumed  stream-line  motion 
but  when  the  velocity  was  above  this  limit  and  the  water  was 
disturbed,  even  though  stream-line  motion  had  existed  before 
the  disturbance,  turbulent  motion  occurred  and  stream-line 
motion  could  not  be  established.  This  limiting  velocity  is  called 
the  lower  critical  velocity. 

The  conditions  of  flow  in  a  pipe  \  in.  in  diameter  are  illustrated 
in  Fig.  91.  The  line  OA  repre- 
sents a  gradual  increase  in 
velocity.  If  the  water  is  not 
disturbed,  stream-line  motion 
will  continue  until  a  velocity 
somewhat  greater  than  3  ft.  per 
second  has  been  reached.  Above 
this  velocity  the  flow  will  always 
be  turbulent.  If  now  the  water, 
starting  with  turbulent  motion, 
is  gradually  decreased  in  velocity 
as  indicated  by  the  line  AB, 
turbulent  motion  will  continue 
until  the  velocity  is  reduced  to 
about  0.5  ft.  per  second.  Below 
this  velocity  stream-line  motion 
will  always  exist. 

In  general,  it  may  be  stated  that  for  any  pipe  carrying  water 
of  a  constant  temperature : 

(a)  There  is  a  certain  velocity  (the  lower  critical  velocity) 
below  which  stream-line  motion  always  exists. 

(6)  There  is  a  certain  velocity  (the  higher  critical  velocity) 
above  which  turbulent  motion  always  exists. 

(c)  Between  the  lower  critical  velocity  and  the  higher  critical 
velocity  the  motion  may  be  either  stream-line  or  turbulent, 
depending  upon  the  initial  condition  of  flow. 

Reynolds  found  that  the  critical  velocity  varied  inversely  as 
the  diameter  and  directly  as  the  viscosity  of  the  water,  the  latter 


138  FLOW  OF  WATER  THROUGH  PIPES 

being  a  function  of  the  temperature,  or  expressed  empirically, 
the  lower  critical  velocity  is 

_0.0388P 
Vl~       d      '  ..........     W 

and  the  higher  critical  velocity  is 
0.246P 


(2) 


d  being  the  diameter  of  the  pipe  in  feet  and 


P  = 


1+0.034  T+Q.QQQ2T2' 


being  a  viscosity  coefficient  in  which  T  is  the  temperature  in 
degrees  Centigrade. 

The  following  are  critical  velocities  in  feet  per  second  obtained 
from  the  above  formulas  for  pipes  of  different  diameters  at  a 
temperature  of  20°  C.  or  68°  F. 

\  in.         1  in.         2  in.         4  in.         6  in.         12  in. 
Lower  ......  0.53        0.26        0.13        0.07        0.04        0.02 

Higher  .....   3.35         1.68        0.84        0.42        0.28        0.14 

As  indicated  by  the  above  table  the  velocities  entering  into 
problems  with  which  the  engineer  has  to  deal  are  ordinarily 
greater  than  the  higher  critical  velocity.  If  not  otherwise  stated, 
therefore,  turbulent  flow  will  be  assumed. 

The  laws  governing  stream-line  motion  are  radically  different 
from  those  governing  turbulent  motion. 

91.  Friction  and  Distribution  of  Velocities.  —  There  is  always 
friction  between  moving  water  and  the  surface  of  the  conduit 
with  which  the  water  comes  in  contact.  If  this  were  not  so  the 
water  in  every  part  of  the  cross-section  would  flow  with  the  same 
velocity.  Fig.  92  shows  the  normal  condition  of  flow  in  a  straight 
pipe  where  there  are  no  disturbing  influences.  Water  particles 
adjacent  to  the  surface  are  retarded  by  friction  and  viscosity 
(Art.  6)  causes  a  retardation  of  the  particles  removed  from  the 
pipe  surface.  The  maximum  velocity  is  at  the  center,  and 
lines  of  equal  velocity  are  concentric  rings  as  shown  in  cross- 


ENERGY  OF  WATER  IN  A  PIPE 


139 


section.  The  velocities  in  any  longitudinal  section  when  plotted 
as  absicssas  with  the  distance  from  one  edge  of  the  pipe  as  ordi- 
nates  approximately  define  an  ellipse.  Experiments  indicate  that 
the  mean  velocity  is  about  0.85  of  the  maximum  velocity.  If  d 
represents  the  diameter,  the  circle  of  mean  velocity  is  approxi- 
mately 0.13d  from  the  surface  of  the  pipe. 


Fig.  92. — Distribution  of  velocities  in  straight  pipe. 

Any  irregularity  or  obstruction  in  a  pipe  or  any  condition 
which  causes  the  water  to  change  its  direction  of  flow  will  change 
the  regular  distribution  of  velocities.  A  bend  in  a  pipe,  for 
example,  causes  the  line  of  maximum  velocity  to  move  from 
the  axis  of  the  pipe  towards  its  concave  side.  Fig,  93  shows  the 


Fig.  93. — Distribution  of  velocities  in  curved  pipe. 

actual  distribution  of  velocities  in  a  curved  pipe  from  measure- 
ments by  Saph  and  Schoder. 

92.  Energy  of  Water  in  a  Pipe. — The  energy  contained  in  a 
stream  of  water  assumed  to  be  moving  with  a  uniform  velocity, 
that  is,  with  the  same  velocity  in  every  part  of  its  cross-section, 
is  given  by  the  formula, 


KE=W^-, 


(4) 


140  FLOW  OF  WATER  THROUGH  PIPES 

W  being  the  weight  of  water  which  moves  past  a  cross-section  in 
one  second  with  a  uniform  velocity  v.  In  the  previous  chapter, 
Art.  72,  it  has  been  shown  that  for  the  same  average  velocity, 
the  energy  of  moving  water  in  an  open  channel  is  greater 
for  non-uniform  velocity  in  a  cross-section  than  for  uniform 
velocity.  The  same  is  manifestly  true  for  pipes,  and  since,  as 
explained  in  the  preceding  article  the  velocity  in  pipes  is  never 
uniform,  the  kinetic  energy  of  water  in  a  pipe  is  given  by  the 
formula, 

KE=<*w£g,      .......    (5) 

in  which  a,  a  coefficient  depending  for  its  value  upon  the  distri- 
bution of  velocities  in  the  pipe,  is  always  greater  than  unity. 
Experiments  by  Bazin  and  others  indicate  that  for  a  straight 
pipe,  a  has  a  mean  value  of  about  1.06. 

In  problems  involving  the  flow  of  water  in  pipes  it  is  common 
to  assume  that  the  velocities  at  all  points  of  a  cross-section  are 
equal,  or  that  a  equals  unity  and  therefore,  the  kinetic  energy 

v2 
contained  in  1  Ib.  of  water  (or  the  velocity  head)  is  equal  to  ^-. 

Bernoulli's  equation,  when  written  between  two  points  in  a  fila- 
ment, then  applies  to  the  entire  cross-section  in  which  the  points 
lie.  The  error  introduced  by  assuming  a  equal  to  unity  is  not 
usually  of  serious  consequence. 

93.  Continuity  of  Flow  in  Pipes. — In  any  pipe  flowing  full, 
within  the  limits  of  error  resulting  from  the  assumptions  that 
water  is  incompressible   and  the   pipe   inelastic,   at  any  given 
instant  the  same  quantity  of  water  is  passing  every  cross-section 
of  the  pipe.     This   statement  implies   continuity   of   flow    (see 
Art.   41)    and  holds   true   even   when   the   flow   is   unsteady,  a 
condition  which  exists  when  the   head   producing  discharge  is 
variable. 

94.  Loss   of  Head. — If   there   were   no   friction   losses,    the 
velocity  at  which  water  would  discharge  from  a  pipe,  Figs.  94 
and  95,  would  be  vt=^2gH,  the  same  as  for  an  orifice.     For 
a  horizontal  pipe  of  uniform  diameter,  Fig.  94,  there  would  be  no 
pressure   other  than  that  resulting  from  the  weight  of  water 
within  the  pipe  and  water  would  not  rise  in  the  piezometer  tubes 
at  m  and  n.     In  any  long  pipe  or  system  of  pipes,  however,  by 


LOSS  OF  HEAD 


141 


far  the  greater  portion  of  the  total  head,  H,  is  used  in  overcoming 
friction. 

If  there  is  no  change  in  the  diameter  of  a  pipe,  the  difference 
in  height  of  water  columns  in  piezometer  tubes  at  any  two 
sections  measures  the  loss  of  head  due  to  friction  between  those 
sections.  In  Fig.  94  the  loss  of  head  between  sections  at  ra  and 
n  is  hm—hn.  In  Fig.  95,  which  represents  a  system  of  pipes  of 
different  diameters,  hi—hm  is  the  loss  of  head  between  sections 
at  I  and  m  plus  the  increase  in  velocity  head  at  ra  over  that  at  I. 


Fig.  94. — Pipe  discharging  from  reservoir. 

Similarly,  hm—hn  (Fig.  95)  is  the  loss  of  head  between  sections  at 
ra  and  n  minus  the  decrease  in  velocity  head. 

Considering  the  system  of  pipes  illustrated  in  Fig.  95,  Ber- 
noulli's equation  may  be  written  between  a  point  S  in  the  water 
surface  and  another  point  E  at  the  outlet  as  follows: 


(6) 


HI  being  the  total  loss  of  head  from  all  causes  and  the  remainder 
of  the  nomenclature  being  as  indicated  in  the  figure.  Since  Vs 
may  be  considered  as  equal  to  zero  and  PS=PE= atmospheric 
pressure,  equation  (6)  reduces  to 

Vr2 
rr  rr  VE      |     TJ  /I-T\ 

ZS~ZE~^  Hl'  • 
or  since  ZS  —  ZE=H,  the  total  head 


(8) 


142 


FLOW  OF  WATER  THROUGH   PIPES 


This  means,  that  for  a  pipe  discharging  into  the  air,  the 
total  head  is  equal  to  the  velocity  head  at  the  end  of  the  pipe 
plus  the  sum  of  all  friction  losses.  Since  the  velocity  head  at 
exit  must  be  provided  out  of  the  total  head,  H,  it  is  usually  con- 
sidered in  the  same  manner  as  lost  head.  It  should  be  remem- 
bered, however,  that  as  the  water  leaves  the  pipe  it  still  retains 
the  energy  represented  by  its  velocity  head. 

In  the  case  of  a  pipe  connecting  two  reservoirs,  Fig.  97,  the 
water  in  the  upper  reservoir  has  a  velocity  of  zero  and  it  finally 
comes  to  rest  in  the  lower  reservoir.  The  reservoirs  may  be 
considered  as  parts  of  the  pipe  system  in  which  the  velocities 


Fig.  95.  —  Pipe  of  more  than  one  diameter. 

are  zero,  the  entire  head,  H,  being  utilized  in  overcoming  friction  ; 
whence 


Frictional  losses  result  from  various  causes.  In  any  pipe  in 
which  the  diameter  remains  unchanged  and  there  are  no  con- 
ditions tending  to  disturb  a  regular  distribution  of  velocities,  the 
only  loss  of  head  is  that  due  to  the  combined  effects  of  viscosity 
and  friction  between  the  moving  water  and  the  surface  of  the 
pipe.  This  loss  of  head  is  commonly  referred  to  as  loss  of  head  due 
to  friction.  Other  losses  of  head  are  those  which  result  from 
changing  the  velocity  or  direction  of  flow. 

In  ordinary  pipe  lines  the  loss  of  head  due  to  friction  is  the 
greater  portion  of  the  total  head.  Frequently  all  other  losses 
are  so  small  in  comparison  as  to  be  negligible.  Cases  arise, 
however,  which  require  careful  consideration  of  these  losses  and 


LOSS  OF   HEAD  143 

serious  errors  may  result  from  neglecting  them.  Losses  of  head 
other  than  the  loss  of  head  due  to  friction  are  commonly  known 
as  minor  losses. 

The  following  are  the  principal  causes  of  loss  of  head  in 
pipes,  together  with  the  symbols  which  will  be  used  to  designate 
these  losses.  (All  losses  except  (a)  are  minor  losses.) 

(a)  A  continuous  loss  of  head  due  to  friction  between  the 
moving  water  and  the  inner  surface  of  the  pipe,  and  to  viscosity. 
This  loss  is  commonly  referred  to  as  the  loss  of  head  due  to  friction, 
and  is  designated  by  the  symbol  /i/. 

(6)  A  loss  of  head  at  the  entrance  to  a  pipe,  ho,  the  loss  occurring 
where  the  very  low  velocity  in  the  reservoir  (usually  considered 
zero  velocity)  changes  to  the  velocity  in  the  pipe.  This  is  called 
the  loss  of  head  at  entrance. 

(c)  A  loss  of  head,  ha,  which  occurs  where  a  pipe  discharges 
into  a  reservoir  or  other  body  of  comparatively  still  water.     This 
will  be  called  the  loss  of  head  at  discharge. 

(d)  A  loss  of  head,  hc,  at  the  place  where  a  pipe  changes  to  a 
smaller  diameter  thus  causing  an  increase  in  velocity .    This  is  called 
loss  of  head  due  to  sudden  or  gradual  contraction,  depending  upon 
whether  the  contraction  takes  place  abruptly  or  by  means  of  a 
tapered  connection  between  the  two  pipes  allowing  the  change 
in  velocities  to  be  made  gradually.     The  loss  of  head  at  entrance 
(referred  to  under  (6)  above)  is  evidently  a  special  case  of  loss 
due  to  contraction. 

(e)  A  loss  of  head,  he,  at  the  place  where  a  pipe  changes  to  a 
larger  diameter  thus  causing  a  decrease  in  velocity.     This  is 
called  loss  of  head  due  to  sudden  or  gradual  enlargement,  depend- 
ing upon  whether  the  enlargement  takes  place  abruptly  or  by 
means  of  a  tapered  connection   between  the  two  pipes  allowing 
the   change  in  velocities  to  be  made  gradually.     The  loss  of 
head  at  discharge  (referred  to  under  (c)  above)  is  evidently  a 
special  case  of  loss  of  head  due  to  enlargement. 

(/)  A  loss  of  head,  hg,  caused  by  obstructions  in  a  pipe  line, 
such  as  gates  or  valves.  Obstructions  cause  the  water  to  pass 
through  a  restricted  area  for  a  short  distance,  thus  causing  first 
a  sudden  increase  in  velocity  and  then  a  sudden  return  to  the 
original  velocity.  This  will  be  called  the  loss  of  head  due  to 
obstructions. 

(g)  A  loss  of  head,  hi>,  at  bends  or  curves  in  pipes,  in  addition 


144 


FLOW  OF  WATER  THROUGH  PIPES 


to  the  loss  which  occurs  in  an  equal  length  of  straight  pipe. 
This  is  designated  the  loss  of  head  due  to  bends. 

If  the  symbol  HI  is  used  to  designate  all  losses  of  head  in  a 
pipe  line,  the  loss  of  head  due  to  friction  being  represented  by 
hf  and  all  minor  losses  by  #2, 


(9) 


in  which 


95.  Hydraulic  Gradient. — The  locus  of  the  elevations  to  which 
water  will  rise  in  a  series  of  piezometer  tubes  inserted  in  a  pipe 
line  is  called  the  hydraulic  gradient  or  hydraulic  grade  line. 
The  hydraulic  gradient  of  a  straight  pipe  of  uniform  diameter 
having  the  same  degree  of  roughness  of  interior  surface  through- 
out is  a  straight  line.  In  Fig.  94  the  line  ac  is  the  hydraulic 
gradient  for  the  pipe.  Where  a  pipe  changes  in  diameter  or 
where  for  any  reason  there  is  a  change  in  velocity  or  direction 
of  flow,  there  is  a  break  in  the  hydraulic  gradient,  the  change 
in  elevation  being  the  combined  effects  of  the  change  in  velocity 
head,  where  velocity  changes  occur,  and  the  loss  of  head  due 
to  friction  or  turbulence.  The  broken  line  aiCte&ifoc^dife, 
Fig.  95,  represents  the  hydraulic  gradient  for  the  system  of  pipes 
shown.  The  hydraulic  grade  line  thus  indicates  all  losses  of  head 
and  changes  in  velocity  head. 


FIG.  96. 

96.  Loss  of  Head  Due  to  Friction  in  Pipes. — Fig.  96  represents 
a  straight  pipe  without  obstructions  or  changes  in  diameter. 
The  loss  of  head,  hf,  in  the  length  I  is  a  measure  of  the  resistance 
to  flow.  The  laws  governing  this  loss  are  intricate  and  are  not 
subject  to  exact  analysis.  There  are,  however,  certain  general  laws 


LOSS  OF  HEAD   DUE  TO  FRICTION   IN   PIPES  145 

which  are  in  the  nature  of  conclusions  resulting  from  observation 
and  experiment,  which  appear  to  govern  fluid  friction  in  pipes 
and  which  are  expressed  in  the  most  generally  accepted  pipe 
formulas  now  in  use.  These  laws  may  be  briefly  stated  as 
follows  : 

(a)  Frictional  resistance  is  independent  of  the  pressure  within 
the  pipe,  and  other  things  being  equal: 

(&)  Friction  between  moving  water  and  the  inner  surface  of 
the  pipe  increases  with  the  roughness  of  the  surface.  This  may 
be  expressed  as  a  coefficient  K  whose  value  increases  with  the 
degree  of  roughness  of  the  pipe. 

(c)  Friction  between  moving  water  and  the  inner  surface  of  a 
pipe  is  directly  proportional  to  the  area  of  the  wetted  surface; 
that  is,  it  is  proportional  to  the  product  of  the  wetted  perimeter 
and  the  length  or  irdl,  d  being  the  diameter  and  I  the  length  of 
the  pipe. 

(d)  As  the   cross-sectional   area   of  the   pipe   increases,   the 
retarding  influence  of  viscosity  becomes  less,  and  it  usually  is  con- 
sidered to  vary  inversely  as  some  power  of  the  area,  and  there- 
fore of  the  diameter  or  as  1/d*. 

(e)  Frictional  resistance  varies  directly  as  some  power  of  the 
velocity,  or  as  vn. 

(/)  Frictional  resistance  increases  with  the  viscosity  and 
therefore  inversely  with  the  temperature.  This  factor  is  usually 
omitted  from  pipe  formulas,  coefficients  being  selected  which  apply 
to  average  air  temperatures. 

Combining  the  factors  expressed  in  (6),  (c),  (d)  and  (e)  above, 
the  total  head  lost  is  represented  by  the  equation, 

......     (11) 


or  substituting  Kf  and  m  for  KXn  and  x—l  respectively,  the 
general  expression  for  loss  of  head  due  to  friction  in  pipes  may 
be  written, 

(12) 


since     =  s,  formula  (12)  may  be  transposed  to  the  form 


l_ 

n    m   1_ 

r-f-^J   ff, 


146  FLOW  OF  WATER  THROUGH   PIPES 


or 


1 

substituting  K"  for  ( -^r. )   ,     y  f  or  -  ,   and  z  for  -, 

\A  /  n  n 


v  =  K"dvsz,     .........     (13) 

or  since  the  hydraulic  radius  r,  for  a  circular  pipe  flowing  full, 
equals  d/4,  or  d=4r,  formula  (13)  may  be  written, 


or  substituting  K'"  for  4?K", 

v  =  K"W  ..........     (14) 

Each  of  the  above  formulas,  though  expressed  differently,  con- 
tains all  of  the  factors,  excepting  temperature,  which  are  believed 
to  affect  fluid  friction.  The  base  formulas  for  friction  losses  in 
pipes  are  commonly  written  in  any  of  the  three  forms  expressed 
by  equations  (12),  (13),  and  (14). 

The  further  consideration  of  loss  of  head  due  to  friction  in 
pipes  must  be  purely  empirical.  The  values  of  coefficients  and 
exponents  to  be  applied  to  the  base  formulas  are  determined 
from  the  available  experimental  data.  Of  the  large  number  of 
published  formulas  for  determining  the  loss  of  head  due  to  friction 
in  pipes,  only  a  few  are  given. 

It  should  be  kept  in  mind  that  in  all  of  the  following  formulas, 
hf,  Z,  d  and  other  linear  quantities  must  be  expressed  in  feet 
and  v  must  be  expressed  in  feet  per  second. 

97.  The  Chezy  Formula.  —  This  formula  deserves  a  place  of 
prominence  among  pipe  formulas  not  only  because  it  represents 
the  first  successful  attempt  to  express  friction  losses  in  algebraic 
terms,  but  also  because  it  embodies  all  of  the  laws  of  fluid  friction 
as  they  are  understood  and  applied  at  the  present  time,  and  with 
certain  modifying  factors  that  have  been  found  necessary,  its 
use  is  now  more  general  than  that  of  any  other  formula  either 
for  pipes  or  open  channels. 

As  written  by  Chezy  in  1775  this  formula  is 


(15) 
=  — 
is  the  rate  of  slope  of  the  hydraulic  gradient.     It  will  be  observed 


in  which  v  is  the  mean  velocity,  r  is  the  hydraulic  radius  and  s  =  — 


THE  CHEZY  FORMULA  147 

that  formula  (15)  is  of  the  same  form  as  (14),  y  and  z  being  each 
\  and  C  being  substituted  for  K'n '. 

The  coefficient  C  was  supposed  by  Chezy  to  be  constant, 
but  it  is  now  known  to  vary  with  the  degree  of  roughness  of 
the  surface  with  which  the  water  comes  in  contact  as  well  as 
with  the  velocity  and  hydraulic  radius  (or  diameter).  Since  C 
appears  to  be  a  function  of  v  and  r  the  Chezy  formula  evidently 
does  not  accurately  express  the  law  of  fluid  friction.  In  the  ideal 
formula,  the  coefficient  would  vary  only  with  the  roughness  of 
the  channel,  and  many  attempts  have  been  made  to  obtain  a 
formula  with  such  a  coefficient  expressing  v  as  a  function  of  r 
and  s.  These  attempts  have  met  with  rather  indifferent  success. 

Formula  (15)  is  used  with  an  accompanying  table  giving 
values  of  C  for  different  velocities,  diameters  and  kinds  of  pipe. 
The  table  on  page  148  gives  approximate  average  values  of  C  for 
four  different  kinds  of  pipe,  as  obtained  from  the  available  experi- 
mental data. 

In  an  account  of  experiments  on  the  flow  of  water  in  pipes, 
published  by  Darcy  l  in  1857,  he  expressed  the  Chezy  formula 
in  the  form, 


the  relations  between  C  and  /  in  formulas  (15)  and  (16)  being 
/=77o     and     C  =  2-\  -r. 


It  will  be  observed  that  formula  (16)  may  be  obtained  from 
formula  (12)  by  writing  n  =  2  and  w=l,  the  two  formulas  being 
of  the  same  general  form. 

From  his  experiments  Darcy  deduced  the  following  values  of 
/,  as  representing  the  mean  of  his  observations. 

For  new,  clean  cast-iron  pipes,  d  being  the  diameter  of  the 
pipe  in  feet, 


For  old  cast-iron  pipes, 


1  M.  H.  DARCY:    Recherches  Experimentales  Relatives  au  Mouvement 
de  1'eau  dans  les  Tuyaux.     Paris,  1857. 


148 


THE   CHEZY   FORMULA 


p 
g 

H   O 

&  w 

PHCE 


s§ 
°l 


CO  CO  O  O5  to         OO  03  CO  OS  iH         CO  t~—  O  CO  to 


00  to  Os  C^l  to        l>-  TH  -*HH  CO  OS 
t^OOOOOSOS        OSOOOO        T-HTHTH 


to  <M  CO  OS  C^        HH  b-  O  <N  »O        b-  i— i  HH  l^  OS        TH  10  OO  O  <M 
t^OOGOOOOS        OSOSOOO        OI-HTHTHTH        C^  <N  C<l  CO  CO 


:OSC^H/II>.        OSCOtOOOO        C^tOoO 
00  OS  OS  OS       OS  O  O  O  TH       TH  I-H  TH 


II 

O  fq 


8 


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OSOTHrHTH          C^C^fNCOCO          CO^^tOlO         .tOCOCOCOt^- 


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GO  OS  OS  OS  O   O  O  TH  TH  TH   <M  C^  C5  CO  CO   CO  CO  HH  HH  HH 


1>OOOOOOO5 


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cocot^t^t^-   i>  oo  oo  oo 


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osososos     ooooo 


OOGO  GO  00        COOS  OS 


CO  GO 

OS  OS 


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t^oooooooo      ooosososi 


CO  to   l>-  O  <M  HH  CO    GO  TH  HH  cO  GO 

1>  00  GO  00  00   00  OS  OS  OS  OS 


TH    T-H   T-H    1—  1   T-H 


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OS  OS  O  O   OTHTHTHTH 


GO  <M  10  OOO 

CO  Tjn  Tt  -rH  >O 


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00  00  OS  OS  OS   O  O  O  TH  TH   TH  <M  <N  C^l  <N    CO  CO  CO  HH  HH 


i-o.ll 

Sfl 


co  OOO<M  ia 


FLOW  OF  WATER  THROUGH   PIPES 


149 


§8 


§1 

Qfe 


CO  CO 


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loo  ooooo  ooooo 


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ooooo 


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ooooo 


!>.  OS  lO  CO  TH 


c 

ooooo 


OOt^CO  UTiiO 

ooooo 


lO  T-H  GO  CO 

"*  ^  CO  CO 

o  o  o  o 


^  (M  O  O5t^ 
CO  CO  CO  <M  <N 

OOOOO 


CO  ^  CO  <M  T-I 
C^  <M  <M  C^l  <M 

ooooo 


O  O5  C75  00  t^ 
(M  i—  I  i—  (  i—  I  i—  I 

ooooo 


Q  S 

Is 


o| 


8 


111 


00  t"~  CO  *O  rtl 

ooooo 


•^  CO  CM  i—i  i— I 

ooooo 


ooo 


§§§§§ 


O5  00  t^  CO 

0000 


ooooo 


Cq  ^  rH  OO 

ooooo 


CO  CO(M  (M  (M 


C^  C^J  i—  I 


ooooo 


^  COCOCN  (N 

ooooo 


t^  O5  lO  CO  i—  1 

o'go'o'o3 


CO  i—  1  O  O5  00 

SS8SS 


00  t^.  CO  »O  iO 

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t>.  »O  >O  iO  "^ 


11 


IO(N  OCXDCO 

SSS88 


CO  CO  CO  <N  <M 


t>.  lO 


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t^  lO  >O  lO  ^t1 

ooooo 


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•^  ^  -*1  CO  CO 
OOOOO 


••*  CO  CO 
(M  (M  (M 


o 

88 


01  oot^co  10 
ooooo 


ooooo      ooooo 


"*  00  »O  rt<  (N 
CO  <N  (N  C^  C^ 

ooooo 


ooooo 


CO  iO^Tf<  CO 

ooooo 


CO  <M  T-l  i-H  i-H 

ooooo 


lO  O  1>-  >O  "^ 
CO  CO  OQ  O5  O5 

ooooo 


co  I-H  o  01  oo 

<M  C^  <M  T— i  i— I 

ooooo 


ooooo 


COCO(M  <N  r-t 
OOOOO 


O5  OOI>-  CO  »O 

ooooo 


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•£    oT  02 

1^1 

S^M 


OOTtno 
f-4  C<l  CO 


150 


FLOW  OF  WATER  THROUGH  PIPES 


Later  experiments  have  indicated  that  /  (similar  to  C  in 
formula  (15)  )  varies  with  v  as  well  as  with  d.  Formula  (16), 
the  same  as  formula  (15),  is  used  more  satisfactorily  with  an 
accompanying  table  of  coefficients.  The  table  on  page  149  gives 
average  values  of  /  for  different  kinds  of  pipe.  The  two  formulas, 
(15)  and  (16),  used  in  connection  with  the  coefficients  tabulated 
on  pages  148  and  149,  respectively,  will  give  the  same  results. 
The  formula  is  written  in  the  two  forms  merely  as  a  convenience 
in  solving  different  types  of  problems. 

The  values  of  C  and  /  given  in  the  tables  on  pages  148  and  149 
refer  not  only  to  pipes  of  the  particular  materials  listed  but  to 
any  pipes  of  similar  degrees  of  roughness.  The  problem  of 
selecting  the  proper  coefficient  for  a  given  condition  is  one  with 
which  the  engineer  is  continually  confronted,  and  in  making  such  a 
selection  experience  is  the  best  teacher.  It  is  important  to.  know 
the  most  probable  value  of  a  coefficient  and  the  maximum  per 
cent  of  error  likely  to  result  from  its  use.  The  average  values  of 
C  and  /  listed  in  the  tables  may  give  results  in  error  as  much  as 
20  per  cent  plus  or  minus. 

The  following  values  of  /  for  2J-in.  fire  hose  are  given  by 
Freeman : 

VALUES  OF  /  IN  CHEZY  FORMULA  FOR  2|-iN.  FIRE  HOSE 


Description. 

VELOCITY  IN  FEET  PER  SECOND. 

4 

6 

10 

15 

20 

Unlined  canvas                     .... 

.038 
.032 
.024 

.038 
.031 
.023 

.037 
.031 
.022 

.035 
.030 
.019 

.034 
.029 
.018 

Rough  rubber-lined  cotton 

Smooth  rubber-lined  cotton  

98.  Hazen- Williams  Formula. — This  formula  which  is  of  the 
form  of  (14)  is 

y=Cir0-63s°-540.001-0-04 (17) 

It  is  designed  for  both  pipes  and  open  channels,  but  is  used  more 
commonly  in  connection  with  pipes.  The  selection  of  exponents 
was  made  with  a  view  to  obtaining  a  minimum  variation  in  Ci 
for  all  conduits  of  the  same  degree  of  roughness.  In  other 


HAZEN-WILLIAMS   FORMULA  151 

words  the  aim  was  to  select  values  for  exponents  such  that  Ci 
would  be,  as  nearly  as  practicable,  a  function  only  of  the  degree 
of  roughness  of  the  channel  and  not  of  r  and  s.  The  following  is 
written  by  the  authors  1  of  the  formula. 

"  If  exponents  could  be  selected  agreeing  perfectly  with  the 
facts,  the  value  of  Ci  would  depend  upon  the  roughness  only, 
and  for  any  given  degree  of  roughness  Ci  would  then  be  a  constant. 
It  is  not  possible  to  reach  this  actually,  because  the  values  of  the 
exponents  vary  with  different  surfaces,  and  also  their  values  may 
not  be  exactly  the  same  for  large  diameters  and  for  small  ones, 
nor  for  steep  slopes  and  for  flat  ones.  Exponents  can  be  selected, 
however,  representing  approximately  average  conditions,  so  that 
the  value  of  Ci  for  a  given  condition  of  surface  will  vary  so  little 
as  to  be  practically  constant.  Several  such  '  exponential  ' 
formulas  have  been  suggested.  These  formulas  are  among  the 
most  satisfactory  yet  devised,  but  their  use  has  been  limited  by 
the  difficulty  in  making  computations  by  them.  This  difficulty 
was  eliminated  by  the  use  of  a  slide-rule  constructed  for  that 
purpose. 

"  The  exponents  in  the  formula  used  were  selected  as  repre- 
senting as  nearly  as  possible  average  conditions,  as  deduced  from 
the  best  available  records  of  experiments  upon  the  flow  of  water 
in  such  pipes  and  channels  as  most  frequently  occur  in  water- 
works practice.  The  last  term,  0.001  ~('04,  is  a  constant,  and 
is  introduced  simply  to  equalize  the  value  of  C\  with  the  value 
in  the  Chezy  formula,  and  other  exponential  formulas  which  may 
be  used,  at  a  slope  of  0.001  instead  of  at  a  slope  of  1." 

Since  0.001  ~°'°4=  1.318,  the  formula  may  be  written, 

-6V-54  .......     (18) 


The  authors  of  the  formula  give  the  following  values  of  C\  for 
pipes: 

For  extremely  smooth  and  straight  pipes  .  .  .  .  C\  —  140 

For  very  smooth  pipes  ....................  Ci  =  130 

For  new  riveted  steel  pipes  ................  C\  =  110 

For  estimating  discharges  of  pipe  lines  where  the  carrying 
capacity  after  a  series  of  years  is  the  controlling  factor,  values  of 

1  WILLIAMS  AND  HAZEN:  Hydraulic  Tables.     Third  Edition,  1920. 


152  FLOW  OF  WATER  THROUGH  PIPES 

Ci  =  100  for  cast-iron  pipe  and  Ci  =  95  for  riveted  steel  are 
recommended.  For  the  smaller  sizes  of  pipes  a  somewhat  lower 
value  of  Ci  should  be  used. 

For  smooth  wooden  pipes  or  wooden-stave  pipes,  C\  =  120. 

For  vitrified  pipes,  Ci  =  110. 

For  old  iron  pipes  in  bad  condition,  Ci  =  80  to  60,  and  for 
small  pipes  badly  tuberculated,  C\  may  be  as  low  as  40. 

99.  King  Formula.  —  Consider  the  base  formula  (12),  which 
for  convenience  of  reference  is  here  repeated, 


(12) 


It  has  been  found  from  experiments  on  a  great  many  kinds  and 
sizes  of  pipes  that  no  value  of  n  can  be  found  which  does  not 
vary  under  different  conditions  of  flow.  The  extreme  range  of 
variation,  from  investigations  by  Lea  1  and  others,  is  from  about 
1.75  to  2.08.  On  the  other  hand,  it  appears  that  a  mean  value 
of  m  of  1.25  may  be  assumed  without  introducing  any  serious 
inconsistencies.  Formula  (12)  has  therefore  been  modified  by 
Lea  to  the  form, 


KI  and  n  each  being  given  variable  values  depending  upon  the 
degree  of  roughness  of  the  pipe. 

The  formula  also  may  be  written, 

h'=Kd^£g  ........     (20) 

in  which 


Formula  (20)  expresses  the  loss  of  head  due  to  friction  as  a 
function  of  the  velocity  head.  This  is  sometimes  convenient 
since  miner  losses  are  usually  thus  expressed  (Art.  102).  The 
formula  is  somewhat  simpler  to  use  than  a  formula  in  which  v 
has  a  fractional  exponent.  On  the  other  hand,  since  K  varies 

1  F.  C.  LEA:  Hydraulics,  p.  139.  H.  W.  KING:  Handbook  of  Hydraulics, 
T>.  159. 


KING  FORMULA 


153 


only  with  v  and  not  with  d,  a  much  simpler  table  of  coefficients 
is  required  for  formula  (20)  than  is  required  for  the  Chezy  formula. 
Average  values  of  K  to  be  used  with  formula  (20)  are  given  in 
the  table  below.  These  values  are  the  same  as  the  values  of  / 
on  page  149  for  d  — 12  inches. 

I     v2 
VALUES  OF  K  IN  THE  FORMULA  hf=K    125  — 


Velocity, 

Clean 

Old 

Clean 

Clean 

in 

cast-iron 

cast-iron 

wooden 

concrete 

feet  per  second 

pipe 

pipe 

pipe 

pipe 

2 

.021 

.038 

.025 

.029 

5 

.019 

.038 

.020 

.025 

10 

.018 

.038 

.017 

.023 

20 

.016 

.038 

.015 

.021 

Formula  (20)  may  also  be  transposed  and  written, 


).625 


in  which  hf/l  =  s  is  the  slope  of  the  hydraulic   gradient.     Sub- 
stituting this  value  and  writing  C'  for  A/-J, 

v=Cfsy^-Q25.  (21) 


In  this  form  the  formula  is  more  convenient  for  certain  types 
of  problems.  Average  values  of  C'  are  given  in  the  following 
table : 

VALUES  OF  C'  IN  THE  FORMULA  v=C'slAd°-625 


Velocity, 

Clean 

Old 

Clean 

Clean 

in 

cast-iron 

cast-iron 

wooden 

concrete 

feet  per  second 

pipe 

pipe 

pipe 

pipe 

2 

55 

42 

51 

47 

5 

58 

42 

57 

51 

10 

61 

42 

62 

53 

20 

63 

42 

66 

56; 

154  FLOW  OF   WATER  THROUGH  PIPES 

Formulas   (15),   (16),   (20)   and   (21),   when  used  with  their 
accompanying  tables  of  coefficients,  all  give  the  same  results. 

100.  General  Discussion  of  Pipe  Formulas.  —  The  foregoing 
formulas  represent  the  more  common  types  of  formulas  for  deter- 
mining the  loss  of  head  due  to  friction  in  pipes.     There  are  an 
indefinite  number  of  formulas,   many  of  which  possess  merit. 
The  choice  of  one  formula  over  another  is  not  of  so  great  impor- 
tance as  the  careful  and  intelligent  use  of  the  formula  after  it  is 
selected.     The  engineer  should  select  the  formula  for  general 
use  which  he  believes  to  be  in  the  most  convenient  form,  and 
after  adopting  it  he  should  endeavor  to  become  familiar  with  its 
coefficients.     The  tables  contained  in  this  volume  are  sufficient 
for    class   room   exercises,    but   the   practicing   engineer   should 
extend  his  knowledge  of  coefficients  by  study  and  observation, 
and    obtain    values    from    actual    measurements    whenever    the 
opportunity  offers. 

101.  Friction  Formula    for    Non-turbulent    Flow.  —  If   v'   is 
the  velocity  in  feet  per  second,  d\  the  diameter  of  the  pipe  in 
inches,  hf  the  friction  loss  in  a  length  /,  and  P  the  viscosity 
coefficient  (formula  (3),  Art.  90),  the  velocity  in  a  pipe  where 
stream-line  flow  exists  according  to  Reynolds  is 

.......     (22) 


If  a  case  be  assumed  where  dt=l  in.,  hf=  1  ft.  and  1=  100  ft., 
v'  for  a  temperature  of  zero  degrees  Centigrade  is  3.61  ft.  per 
second,  and  for  higher  temperatures  the  velocity  would  be  greater. 
The  table  on  page  138  shows  this  velocity  to  be  above  the  higher 
critical  velocity  and  the  flow  must  be  turbulent.  Formula  (22) 
therefore  does  not  apply  and  one  of  the  formulas  for  turbulent 
flow  should  be  used. 

102.  Detailed  Study  of  Hydraulic  Gradient  and  Minor  Losses. 
—  In  the  discussion  of  loss  of  head  due  to  friction  (Art.  96),  it 
has  been  shown  that,  other  things  being  equal,  the  loss  of  head 
varies  as  vn  and  that  usually  n  is  less  than  2  but  does  not  vary 
greatly  from  this  value.  In  some  formulas,  therefore,  this  loss 
of  head  is  expressed  as  a  function  of  the  velocity  head  and  coeffi- 
cients varying  in  value  with  v  are  applied  to  the  formulas  to  make 
them  represent  average  friction  losses  as  given  by  experiments. 

In  a  similar  manner  it  has  been  found  that  minor  losses  (Art. 


DETAILED  STUDY  OF  HYDRAULIC  GRADIENT         155 

94)  vary  roughly  with  the  square  of  the  velocity  and  they  are 
commonly  expressed  in  formulas  as  functions  of  the  velocity 
head.  Variable  coefficients  are  then  applied  to  these  formulas 
so  as  to  make  them  give  losses  in  accordance  with  the  available 
experimental  data.  These  losses  (see  Art.  94)  expressed  alge- 
braically are 

V2  V2  V2 

ho  =  K0^-,    hd  =  Kd-^}     hc  =  Kc—,     etc., 
and  formula  (10),  page  144,  may  be  written, 


+K-  •  (23) 

In  the  above  equation,  v  is  a  general  expression  for  velocity. 
It  is  the  velocity  in  the  pipe  where  the  loss  of  head  occurs  and 
in  case  of  enlargement  or  contraction  it  is  the  velocity  in  the 
smaller  pipe.  KQ,  Kd,  Kc,  etc.,  are  variable  coefficients  whose 
values  must  be  determined  from  experiments. 

Fig.  97  illustrates  entrance  and  discharge  conditions  for  a 
pipe  leading  from  one  reservoir  into  another  reservoir  at  a  lower 
elevation.  The  water  starts  with  zero  velocity  in  the  upper 
reservoir,  finally  coming  to  rest  in  the  lower  reservoir,  and  all 
of  the  energy  represented  by  the  difference  in  elevation  of  water 
surfaces  is  utilized  in  overcoming  resistance. 

In  Fig.  97,  a\a%  represents  the  hydraulic  grade  line  which 
results  from  changing  the  velocity  of  the  water  from  zero  to  the 
velocity  which  it  attains  in  the  pipe.  The  vertical  distance 
between  a\  and  #2,  that  is,  the  distance  which  «2  is  below  the 
surface  of  the  water,  is  the  velocity  head  or  v2/2g,  where  v  is  the 
mean  velocity  in  the  pipe.  The  line  0,10,2,  must  be  considered  as 
the  hydraulic  gradient  of  some  particular  filament  of  water,  such  as 
xy,  since  points  in  other  filaments  which  are  the  same  horizontal 
distance  from  the  entrance  to  the  pipe  may  have  different  veloci- 
ties and  therefore  different  hydraulic  gradients.  It  may  appear 
that  the  pressure  at  any  point  in  the  filament  should  be  that  due 
to  the  weight  of  the  water  column  above  it.  This  would  be  true 
if  the  laws  of  hydrostatics  might  be  applied.  The  laws  of  hydro- 
statics do  not,  however,  apply  to  water  in  motion,  the  pressure 
being  less  than  it  would  be  at  the  same  depth  for  water  at  rest. 
That  this  is  true  has  been  proved  experimentally.  It  also  follows 
from  writing  Bernoulli's  equation  between  a  point  x  where  the 


156 


FLOW  OF  WATER  THROUGH   PIPES 


velocity  is  practically  zero  and  a  point  y  at  the  entrance  to  the 
pipe  where  the  velocity  equals  v,  the  velocity  in  the  pipe.  Assum- 
ing the  points  to  be  of  the  same  elevation  the  equation  becomes 


or  since  vx  is  practically  zero, 


. 


(24) 


The  head  lost  at  entrance  to  a  pipe  takes  place  within  a  distance 
of  about  two  or  three  diameters  from  the  entrance  and  is  similar 
to  the  loss  of  head  in  a  short  tube.  The  line  c^as,  Fig.  97,  is 
the  portion  of  the  hydraulic  gradient  which  shows  this  loss  of 


Fig.  97. — Pipe  connecting  two  reservoirs. 

head.  There  is  a  depression  in  the  hydraulic  gradient  at  o! 
because  of  the  contraction  of  the  jet.  Vertically  below  as,  the 
jet  has  expanded  and  fills  the  tube.  The  head  lost  at  entrance  is 
the  vertical  distance  between  az  and  as,  or  ho. 

Since  the  first  two  or  three  diameters  of  a  pipe  are  similar  to 
a  short  tube,  entrance  losses  for  pipes  may  be  considered  to  be 
the  same  as  for  short  tubes.  The  general  formula  for  loss  of 
head  at  entrance  to  a  pipe  is  then  (formula  (22),  page  81), 


(25) 


in  which  the  coefficient  of  discharge,  C,  depends  for  its  value  upon 
tl^e  conditions  at  entrance,  and  Ko  =  -~—l.     For  convenience  of 


DETAILED  STUDY  OF  HYDRAULIC  GRADIENT 


157 


reference,  values  of  C  and  KQ  given  in  Chapter  VII,  are  repeated 
in  the  following  table : 

COEFFICIENTS  FOR  DETERMINING  Loss  OF  HEAD  AT  ENTRANCE  TO  PIPES 


Entrance  to  pipe 

Reference 

C 

K0 

Inward  projecting              .... 

Art  63 

0  75 

0  78 

Sharp  cornered 

Art  58 

0  82 

0  50 

Slightly  rounded    

Art.  57 

0.90 

0  23 

Bell  mouth                 

Art.  57 

0  98 

0  04 

Since  the  effect  of  entrance  conditions  can  not  be  determined 
accurately  the  selection  of  a  proper  value  of  KQ  is  to  some  extent 
a  matter  of  judgment.  Unless  the  entrance  is  known  to  be  other 
than  sharp  cornered,  a  value  of  0.5  is  commonly  used. 

Conditions  at  the  outlet  of  a  pipe  may  be  illustrated  in  a  similar 
manner.  If  there  were  no  loss  of  head  where  water  enters  the 
lower  reservoir  the  hydraulic  grade  line  would  connect  as  and  e\ 
(Fig.  97),  the  latter  point  being  v2/2g  below  the  surface  of  the 
water.  The  distance  61^2  represents  the  portion  of  the  velocity 
head  lost  through  shock  and  turbulence.  This  may  be  illustrated 
by  writing  Bernoulli's  equation  between  a  point  s  at  the  outlet 
of  the  pipe  where  the  velocity  equals  v,  the  velocity  in  the  pipe, 
and  a  point  u  where  the  velocity  vu  is  practically  zero,  ha  equals 
the  loss  of  head  due  to  turbulence.  If  the  two  points  are  at  the 
same  elevation, 

^-f^  =  >TiM+^+^,         (26) 

or  since  t>«  =  0, 


v2 


(27) 


hu—hs  represents  the  portion  of  the  velocity  head  which  is  not 
lost  but  which  is  reconverted  into  pressure  head.  The  rate  at 
which  this  reconversion  takes  place  is  represented  in  the  figure 
by  the  line  62^3,  the  end  of  the  hydraulic  gradient. 

Expressing  the  head  lost  at  discharge,  ha,  as  a  function  of  the 
velocity  head, 

^=•^ (28) 


158 


FLOW  OF  WATER  THROUGH  PIPES 


The  loss  at  discharge  is  the  special  case  of  loss  of  head  due  to 
sudden  enlargement  in  which  the  ratio  of  smaller  to  larger  dia- 
meter is  practically  zero.  Values  of  Kd  may  therefore  be  taken 
from  column  2  of  the  table  on  page  161.  Since  these  values  are 
nearly  unity  for  the  ordinary  velocities  encountered  in  pipes  it  is 
commonly  considered  that  the  entire  velocity  head  is  lost. 

Change  in  gradient  resulting  from  sudden  contraction  in  pipes 
is  illustrated  in  Fig.  98.  If  there  were  no  loss  of  head  between 
any  two  points,  on  opposite  sides  of  the  contracted  section,  the 
difference  in  heights  of  water  columns  in  two  piezometer  tubes  as 
b  and  e}  above  these  points  would  measure  the  gain  in  velocity 
head.  If  the  two  points  are  considered  so  close  together  that 
pipe  friction  may  be  neglected,  the  difference  in  height  of  water 


Fig.  98. — Sudden  contraction  in  pipe. 

columns  6  and  e  measures  the  gain  in  velocity  head  plus  the  loss 
of  head  due  to  sudden  contraction. 

The  hydraulic  gradient  as  determined  experimentally  is 
illustrated  by  the  line  abcdef.  There  is  a  depression  at  d,  due 
to  contraction  of  the  jet,  similar  to  the  depression  at  a'  in  the 
hydraulic  grade  line  of  Fig.  97.  The  piezometer  tube  c  measures 
the  pressure  in  the  corner  where  there  is  little  or  no  velocity. 
If  piezometer  tubes,  c  and  d,  were  arranged  to  measure  pressures 
near  the  axis  of  the  pipe  where  the  velocities  are  higher,  the 
hydraulic  gradient  would  be  below  bcde  and  would  resemble 
Wd'e. 

It  is  important  to  note  that  the  ordinary  piezometer  tube, 
which  is  set  flush  with  the  inner  surface  of  the  pipe,  measures 
the  pressure  at  the  surface  of  the  pipe  but  does  not  necessarily 
measure  the  pressure  at  points  in  the  same  cross-section  at  some 


DETAILED  STUDY  OF  HYDRAULIC   GRADIENT 


159 


distance  from  the  surface.  In  smooth  straight  pipes  the  differ- 
ence between  pressures  at  the  surface  and  interior  points  is  prob- 
ably not  great  but  the  difference  may  be  quite  large  near  sections 
where  changes  in  diameter  occur. 

The  loss  of  head  due  to  sudden  contraction  expressed  as  a 
function  of  the  velocity  head  is 


(29) 


in  which  Kc  is  an  empirical  coefficient,  and  v  is  the  velocity  in 
the  smaller  pipe.  The  following  table  gives  experimental  values 
of  JC* 

VALUES  OF  THE  COEFFICENT  Kc,  FOR  SUDDEN  CONTRACTION 


Velocity  in 


RATIO  OF  SMALLER  TO  LARGER  DIAMETER 


smaller  pipe, 

V 

0.0 

0.1 

0.2 

0.3 

0.4 

0.5 

0.6 

0.7 

0.8 

0.9 

2 

0.49 

0.49 

0.48 

0.45 

0.42 

0.38 

0.28 

0.18 

0.07 

0.03 

5 

.48 

.48 

.47 

.44 

.41 

.37 

.28 

.18 

.09 

.04 

10 

.47 

.46 

.45 

.43 

.40 

.36 

.28 

.18 

.10 

.04 

20 

.44 

.43 

.42 

.40 

.37 

.33 

.27 

.19 

.11 

.05 

40 

.38 

.36 

.35 

.33 

.31 

.29 

.25 

.20 

.13 

.06 

The  loss  of  head  at  entrance  to  pipes  is  a  special  case  of  loss 
of  head  due  to  contraction.  If  the  body  of  water  is  large  the 
conditions  conform  approximately  to  a  ratio  of  diameters  of 
zero,  and  for -a  square-cornered  entrance,  where  the  end  of  the 
pipe  is  flush  with  a  wall  having  a  plane  surface,  the  values  of  KQ 
are  comparable  with  the  values  of  Kc  in  the  second  column  of  the 
above  table. 

If  the  change  to  a  smaller  diameter  takes  place  gradually,  as 
is  the  case  with  a  gradually  tapering  section  connecting  the  two 
pipes,  or  if  the  corners  of  the  smaller  pipe  are  rounded  so  as  to 
reduce  contractions,  values  of  Kc  will  be  much  smaller  than  those 
given  for  a  sudden  reduction  of  diameter.  If  the  change  is  made 
as  gradually  as  in  a  Venturi  meter  or  if  a  bell-mouth  connection 


160 


FLOW  OF  WATER  THROUGH   PIPES 


(page  85)  between  the  two  pipes  is  used,  Kc  may  become  prac- 
tically negligible. 

Change  in  gradient  resulting  from  sudden  enlargement  is  shown 
in  Fig.  99.  In  this  case  the  change  in  height  of  water  columns 
in  piezometer  tubes  b  and  e,  before  and  after  enlargement, 
measures  the  gain  in  pressure  head  and  this  gain  is  equal  to  the 
loss  in  velocity  head  minus  the  loss  of  head  due  to  sudden 
enlargement,  the  loss  of  head  due  to  pipe  friction  being  assumed 
so  small  as  to  be  negligible.  This  may  be  verified  by  writing 
Bernoulli's  equation  between  points  on  either  side  of  the  enlarge- 
ment. 

The  hydraulic  gradient  as  plotted  from  experiments  by 
Gibson  is  abcde.  The  pressures  shown  by  tubes  c  and  d  being 


2  g 


Fig.  99. — Sudden  enlargement  in  pipe. 

measured  at  the  surface  of  the  pipe  are  undoubtedly  less  than  if 
they  were  measured  near  the  center  of  the  pipe.  In  this  case 
the  piezometer  tube  c  would  read  practically  the  same  as  the 
tube  b  and  the  pressure  would  be  greater  than  that  indicated. 
The  portion  of  the  hydraulic  gradient  bcde  would  then  be  similar 
to  bc'd'e. 

The  loss  of  head  due  to  sudden  enlargement  expressed  as  a 
function  of  the  velocity  head  is 


(30) 


Archer  l  has  shown  from  an  investigation  of  his  own  experi- 


1W.  H.  ARCHER:   Loss  of  Head  Due  to  Enlargements  in  Pipes.     Trans. 
Amer.  Soc.  Civ.  Eng.,  vol.  76,  pp.  999-1026  (1913). 


DETAILED  STUDY  OF  HYDRAULIC  GRADIENT 


161 


ments  and  the  experiments  of  others  that  hc  is  quite  accurately 
represented  by  the    formula, 


A,  =  1.098 


(31) 


in  which  v  is  the  velocity  in  the  smaller  pipe  and  v\  is  the  velocity 
in  the  larger  pipe.  Experiments  at  the  University  of  Michigan 
indicate  that  Archer's  formula  holds  quite  accurately  in  the 
limit  where  v\  is  zero.  In  this  case  the  conditions  become  those 
of  loss  of  head  at  discharge,  discussed  on  page  157. 

By  equating  the  values  of  he  given  by  equations  (30)  and  (31) 
and  transposing,  the  following  value  of  Ke  is  obtained: 


Ke 


1.098 

,,67osi 


d 


1.919 


(32) 


d/di  being  the  ratio  of  the  smaller  to  the  larger  diameter.     The 
following  table  of  values  of  Ke  are  computed  from  formula  (32) : 

VALUES  OF  THE  COEFFCIENT  Ke,  for  SUDDEN  ENLARGEMENT 


Velocity  in 


RATIO  OF  SMALLER  TO  LARGER  DIAMETER 


smaller  pipe, 

V 

0.0 

0.1 

0.2 

0.3 

0.4 

0.5 

0.6 

0.7 

0.8 

0.9 

2 

1.00 

1.00 

0.96 

0.86 

0.74 

0.60 

0.44 

0.29 

0.15 

0.04 

5 

.96 

.95 

.89 

.80 

.69 

,56 

.41 

.27 

.14 

.04 

10 

.93 

.91 

.86 

.77 

.67 

.54 

.40 

.26 

.13 

.04 

20 

.86 

.84 

.80 

.72 

.62 

.50 

.37 

.24 

.12 

.04 

40 

.81 

.80 

.75 

.68 

.58 

.47 

.35 

.22 

.11 

.03 

The  loss  of  head  due  to  enlargement  in  pipes  may  be  reduced 
by  changing  the  diameters  gradually.  If  the  diameter  increases 
at  a  uniform  rate,  the  amount  of  loss  increases  as  the  angle 
between  the  axis  and  surface  of  the  pipe  increases  and  is  practi- 
cally negligible  for  very  small  angles.  Experimental  values  of  Ke 
have  not  been  well  determined  for  gradual  enlargements,  but 
those  given  in  the  following  table  are  the  approximate  mean  of 
such  data  as  are  available.  There  are  not  sufficient  experimental 
data  to  determine  the  extent  to  which  Ke  varies  with  the  velocity. 


162  FLOW  OF  WATER  THROUGH  PIPES 

VALUES  OF  THE  COEFFCIENT  Ke,  FOR  GRADUAL  ENLARGEMENT 


RATIO  OF  SMALLER  TO  LARGER  DIAMETER 

Angle  between  axis 
and  surface  of  pipe 

0.1 

0.2 

0.3 

0.4 

0.5 

0.6 

0.7 

0.8 

0.9 

5° 

0.04 

0.04 

0.04 

0.04 

0.04 

0.04 

0.03 

0.02 

0.01 

15° 

.16 

.16 

.16 

.16 

.16 

.15 

.13 

.10 

.06 

30° 

.49 

.49 

.48 

.48 

.46 

.43 

.37 

.27 

.16 

45° 

.64 

.63 

.63 

.62 

.60 

.55 

.49 

,38 

.20 

60° 

.72 

.72 

.71 

.70 

.67 

.62 

.54 

.43 

.24 

Gates  or  Valves  when  partially  closed  obstruct  the  flow  and 
cause  a  loss  of  head  and  consequent  drop  in  the  hydraulic  gradient. 
The  difference  in  elevation  of  the  hydraulic  gradient  on  opposite 
sides  of  the  obstruction  measures  the  loss  of  head  due  to  the 
obstruction.  Following  the  form  used  for  other  losses,  the  loss 
of  head  in  pipes  due  to  gates,  valves,  or  other  obstructions  may 
be  written, 

lh=K,£,       (33) 

v  being  the  mean  velocity  in  the  pipe.  Experiments  indicate 
that  Kg  does  not  vary  appreciably  with  the  velocity  but  increases 
with  the  amount  of  restriction.  The  following  values  of  Kg  are 
the  average  values  obtained  from  the  best  experimental  data 
available. 

VALUES  OF  THE  COEFFICIENT  Kg,  FOR  OBSTRUCTIONS  IN  PIPES 


Ratio  of  area  of  opening 
to  cross-sectional  area 

K. 

Ratio  of  area  of  opening 
to  cross-sectional  area 

of  pipe 

of  pipe 

0.1 

13.5 

0.5 

2.7 

0.15 

10.0 

0.6 

1.8 

0.2 

8.0 

0.7 

1.1 

0.3 

5.7 

0.8 

0.6 

0.4 

4.0 

0.9 

0.2 

Bends  or  curves  in  pipes  cause  a  gradual  drop  in  the  hydraulic 
gradient  which  is  in  excess  of  the  drop  that  would  occur  in  an 


DETAILED  STUDY  OF  HYDRAULIC  GRADIENT 


163 


equal  length  of  the  same  kind  of  straight  pipe.  From  a  careful 
investigation  of  the  available  experimental  data  Fuller  x  deduced 
the  following  empirical  formula, 

hb  =  cu2-25,       .'•  '.     .   V    .     .'    .     (34) 

hd  being  the  loss  of  head  due  to  bends  in  excess  of  the  loss  which 
would  occur  in  an  equal  length  of  straight  pipe,  v  being  the 
mean  velocity  in  the  pipe  and  c  being  a  coefficient  whose  value 
varies  with  the  radius,  R,  of  the  axis  of  the  pipe  expressed  in 
feet.  Fuller  gives  the  following  values  of  c  for  bends  of  90°: 
For  R  =  Q,c  =  0.0135;  for  R  =  0.5,  c  =  0.0040;  for  R  =  l,  c  =  0.00275; 
for  R  =  3,  c  =  0.0024;  for  #  =  6,  c  =  0.0023;  for  #=10,  c  =  0.00335; 
for  R  =  20,  c  =  0.0060;  for  #  =  30,  c  =  0.0070;  for  #  =  40,  c  =  0.0075; 
for  #  =  60,  c  =  0.0086. 

Following  the  form  used  for  other  losses, 


v2 
hb  =  Kb  — , 


(35) 


and  equating  the  right-hand  members  of  equations  (34)  and  (35) 
and  reducing, 

........     (36) 


From  this  formula  the  values  of  Kb  given  in  the  following  table 
have  been  computed: 

VALUES  OF  THE  COEFFICIENT,  Kb.  FOR  Loss  OF  HEAD  DUE  TO  BENDS  OF  90° 


Mean 

RADIUS  OF  BEND  IN  FEET 

velocity 

in 

pipe,  v 

0 

0.5 

1 

2 

6 

8 

10 

20 

30 

40 

50 

2 

1.03 

0.31 

0.21 

0.19 

0.18 

0.21 

0.26 

0.45 

0.53 

0.57 

0.61 

5 

1.30 

.38 

.26 

.23 

.22 

.26 

.32 

.57 

.67 

.72 

.77 

10 

1.54 

.46 

.31 

.28 

.26 

.31 

.38 

.68 

.79 

.86 

.92 

20 

1.84 

.54 

.37 

.33 

.31 

.37 

.46 

.81 

.95 

1.02 

1.08 

40 

2.18 

.65 

.44 

.39 

.37 

.44 

.54 

.97 

1.12 

1.21 

1.30 

From  the  above  table  it  will  be  seen  that  the  minimum  loss 
of  head  from  bends  occurs  when  the  radius  of  the  axis  of  the 

1  W.  E.  FULLER:  Loss  of  Head  in  Bends.     Journal  of  New  England  Water 
Works  Association,  December,  1913. 


164  FLOW  OF  WATER  THROUGH  PIPES 

pipe  is  from  2  to  6  ft.  For  bends  of  45°  the  coefficients  will  be 
about  three-fourths  and  for  22 J°  about  one-half  of  the  values 
given  in  the  table. 

103.  Part  of  Pipe  Above  Hydraulic  Gradient. — Fig.  100  shows 
a  pipe  of  uniform  diameter  leading  from  a  reservoir  and  discharg- 
ing under  the  head  H.  The  summit,  M ,  is  a  distance  y  above  the 
straight  line  BeC  but  at  a  lower  elevation  than  the  water  surface 
in  the  reservoir.  Two  conditions  will  be  considered:  first, 

Pa~  PC  Pa  —  pv 

where  y< ,  and  second,  where  y> ,  pa  being  atmos- 
pheric pressure  and  p,  being  the  vapor  pressure  corresponding 
to  the  temperature  of  the  water  in  the  pipe. 

Assume  the  pipe  AMSC,  Fig.  100,  to  be  empty  when  water 


Fig.  100. — Part  of  pipe  above  hydraulic  gradient. 

is  turned  into  it  at  A.  Water  will  first  rise  to  the  summit  M 
and  begin  to  flow  down  the  decline  toward  the  depression  S, 
at  which  point  it  will  collect  and  seal  the  pipe  entrapping  air 
between  M  and  S.  Eventually  water  will  discharge  from  the 
outlet  C.  If  the  velocity  is  high  enough  the  air  entrapped 
between  M  and  S  will  be  removed  by  the  flowing  water,  other- 
wise it  will  remain  there  and  obstruct  the  flow.  In  such  cases 
the  air  may  be  removed  by  a  suction  pump  at  the  summit.  If 

there  is  no  air  in  the  pipe  and  y<— °,  assuming    the  loss 

of  head  to  be  uniform,  the  hydraulic  gradient  will  be  the  straight 
line  BeC  and  the  flow  will  be  the  same  as  though  all  of  the  pipe 
were  below  the  hydraulic  gradient. 

ty\       -         /y\ 

If  y>— — —,  the  flow  of  water  will  be  restricted,  even  though 
all  air  is  exhausted  from  the  pipe,  and  the  hydraulic  gradient 


PART  OF  PIPE  ABOVE  HYDRAULIC  GRADIENT         165 

will  no  longer  be  a  straight  line.  This  condition  is  illustrated 
in  Fig.  100.  The  hydraulic  gradient  is  a  straight  line  to  a  point, 

nr\      nr\ 

D,  which  is  a  distance  d  =  — — —  below  the  summit.  M.     At  M 

w 

(assuming  no  air  in  the  pipe)  the  absolute  pressure  in  the  p;pe 
is  the  vapor  pressure  corresponding  to  the  temperature  within 
the  pipe  and  this  pressure  continues  on  down  to  N,  the  pipe 
flowing  partially  full  between  M  and  N.  Throughout  all  por- 
tions of  the  pipe  flowing  full  the  velocity  must  necessarily  be  the 
same  since  the  discharge  is  constant  and  therefore,  assuming  a 
uniform  degree  of  roughness  for  the  pipe,  the  slope  of  hydraulic 
gradient  in  such  portions  must  be  uniform.  In  other  words, 
the  slope  of  EC  must  be  the  same  as  the  slope  of  BD.  Through- 
out the  distance  where  the  pipe  is  not  flowing  full,  the  hydraulic 
gradient,  represented  by  the  line  DE,  is  the  same  vertical  distance, 
d,  below  the  water  surface  in  the  pipe.  The  point  E  is  the 
intersection  of  the  line  CE,  parallel  to  BD,  and  the  line  DE. 
The  section  at  N  where  the  pipe  begins  to  flow  full  is  vertically 
above  E. 

The  conditions  of  flow,  especially  at  low  velocities,  are  not 
usually  as  favorable  as  those  described  above,  because  of  the 
tendency  of  air  to  collect  at  a  summit.  Water  flowing  at  low 
velocities  will  not  remove  air  and  may  even  liberate  it,  and 
cause  air  to  collect  at  high  places  such  as  M,  Fig.  100.  The 
condition  will  be  worse  at  summits  above  the  hydraulic  gradient 
if  the  pipe  leaks,  since  the  movement  of  air  will  be  inward.  In 
such  cases  the  occasional  operation  of  an  air  pump  at  the  summit 
will  be  necessary  to  remove  the  air.  At  a  summit  below  the 
hydraulic  gradient,  where  the  pressure  within  the  pipe  is  greater 
than  atmospheric,  the  air  which  collects  may  be  removed  through 
a  valve. 

Air  at  a  summit  which  is  below  the  elevation  of  the  water 
surface  will  not  stop  the  flow  of  water  entirely  but  will  cause  a 
portion  of  the  pipe  to  flow  partially  full.  Summits  in  pipe  lines 
are  always  objectionable,  and  especially  so  are  summits  above  the 
hydraulic  gradient.  Where  they  cannot  be  avoided  special  pro- 
vision should  be  made  for  removing  the  air  which  collects. 

104.  Special  Problems. — Pipe  lines  may  be  composed  of 
pipes  of  several  diameters  connected  in  series,  or  they  may  branch 
in  different  ways  so  as  to  divide  the  flow,  thus  presenting  a 


166 


FLOW  OF  WATER  THROUGH  PIPES 


variety  of  problems.  Oftentimes  such  problems  may  be  solved 
more  readily  by  trial  solutions,  though  some  formulas  may  be 
derived  which  are  of  assistance.  A  few  special  cases  are  given 
in  the  following  pages.  Problems  of  this  type  are  encountered 
frequently  in  designing  mains  for  city  water  supplies. 

If  the  pipes  are  long  (1000  diameters  or  more)  the  minor 
losses  will  ordinarily  be  comparatively  small  and  are  usually 
neglected.  If,  however,  it  is  desired  to  include  these  losses,  a 
solution  should  be  made  first  neglecting  them  and  then  correcting 
the  results  to  include  them. 

105.  Branching  Pipe  Connecting  Reservoirs  at  Different 
Elevations. — A,  B  and  C  are  three  reservoirs  connected  by 
pipes  1,  2  and  3,  as  shown  in  Fig.  101.  Let  l\,  di,  Qi  and  v\ 


FIG.  101. — Branching  pipe  connecting  three  reservoirs. 

represent,  respectively,  the  length,  diameter,  discharge  and  mean 
velocity  for  pipe  1,  and  the  same  symbols  with  subscripts  2 
and  3,  the  corresponding  terms  for  pipes  2  and  3.  If  a  piezom- 
eter is  assumed  to  be  at  the  junction  P,  the  water  surface  in 
the  tube  will  be  a  certain  distance,  hi}  below  the  surface  in  reser- 
voir A .  The  surface  of  reservoir  B  is  a  distance  HB  =  hi  +7*2  below 
that  of  reservoir  A  and  the  surface  of  reservoir  C  is  Hc=hi-\-h^ 
below  the  surface  of  reservoir  A.  If  hi<HB)  reservoir  A  will 
supply  reservoirs  B  and  C.  If  }II>HB,  reservoirs  A  and  B  will 
supply  reservoir  C.  There  are  many  problems  suggested  by  this 
figure,  in  which  certain  quantities  are  given  with  others  to  be 
determined.  Methods  of  solving  three  of  these  problems  are 
given. 

Case  L — Having  given  the  lengths  and  diameters  of  all  pipes, 
and  elevations  of  the  three  reservoirs;  to  determine  Qi,  Q% 
and  Q3. 


BRANCHING   PIPE   CONNECTING   RESERVOIRS 


167 


This  problem  is  most  conveniently  solved  by  .trial.     Assume 

61  and  solve  for  hi.     Then  using  HB  minus  this  computed  value 
of  h\  for  h2,  the  loss  of  head  due  to  friction  in  pipe  2,  solve  for 
Q2.     Similarly,  using  Hc  minus  the   computed  value  of  hi  for 
friction  loss,  hs,  in  pipe  3,  solve  for  63.     Evidently  61  =  62+63, 

62  being  negative  if  the  direction  of  flow  is  from  B  toward  P. 
The  correct  value  of  Qi  will  lie  between  the  assumed  value  and 
the  computed  value  of  62+63.     Continue  to  assume  new  values 
of   Qi,    between   these   limits,    and   repeat    computations   until 
61  =  02+63. 

It  may  be  found  helpful  in  making  assumptions  to  plot  com- 
puted values  of  61?  Fig.  102, 
against  the  error  made  in  each 
assumption,  that  is,  against  61  ~ 
(62+63).  The  resulting  differ- 
ence may  be  either  plus  or  minus. 
If  the  assumed  values  of  61  are 
well  selected  they  will  define  a 
curve  whose  intersection  with  the 
6i-axis  will  give  the  discharge  as 
accurately  as  is  usually  required. 
The  points  should  be  on  both  sides  of  the  6i-axis  and  preferably 
one  of  the  points  should  be  quite  close  to  it.  Usually  not 
more  than  three  trial  solutions  will  be  necessary. 

This  problem  may  also  be  solved    analytically.      Assuming 
any  formula  for  pipe  friction,  as,  for  example,  the  Chezy  formula, 


10  9  3  7  6 


4  3  2  1  0  1  2  3  4  5^6  7 


FIG.  102. 


910 


h 


also 


and 


From  Fig.  (101), 
and  substituting  the  above  values  of  hi  and  h2, 


(37) 


168  FLOW  OF  WATER  THROUGH   PIPES 

and  in  a  similar  manner, 

TT         f    h    Vl2_Lf    k    ^32.  ,oox 

Hc=fldi2J+f*Z2ii> (38) 

also  since  Qi  =  Q2+Qs 

di2v1=d22V2+d32v3 (39) 

By  solving  equations  (37),  (38)  and  (39)  simultaneously,  vi,  V2 
and  vz  may  be  determined  if  the  other  quantities  are  given. 
Since  fi,  /2  and  /a  are  dependent  upon  vi,  V2  and  v$,  respectively 
for  their  values,  the  equations  must  first  be  solved  with  assumed 
friction  coefficients,  to  be  corrected  after  the  first  solution  for 
velocities  has  been  completed.  With  these  corrected  values  of 
fit  /2  and  /3,  another  solution  of  the  equations  for  more  accurate 
values  of  vi,  v%  and  v%  may  be  made. 

Case  2. — Having  given  the  lengths  and  diameters  of  all  pipes, 
Qi,  and  the  elevations  of  water  surfaces  in  reservoir  A  and  one 
of  the  other  reservoirs  as  B]  to  determine  the  elevation  of  water 
surface  in  reservoir  C. 

Using  Qi,  determine  the  lost  head,  hi,  in  pipe  1.  Then 
h2  =  Hs—hi  is  the  lost  head  in  pipe  2,  using  which,  $2  may  be 
computed.  $2  will  be  plus  or  minus  depending  upon  whether  the 
direction  of  flow  in  pipe  2  is  towards  B  or  P.  Then  Qs  =  Qi  —  $2. 
With  $3  determined,  the  head  lost  in  pipe  3  may  be  computed, 
and  the  elevation  of  water  surface  in  reservoir  C  obtained. 

Case  3. — Having  given  the  lengths  of  all  pipes,  the  elevations 
of  water  surfaces  in  all  reservoirs,  Qi,  and  the  diameters  of  two 
pipes  as  d\  and  d2',  to  determine  cfe. 

Determine  hi,  $2  and  Qs  as  for  Case  2.  Then  with  Qs  and 
hz  =  Hc—hi  known,  compute  d%. 

106.  Compound  Pipe  Connecting  Two  Reservoirs. — The  reser- 
voirs A  and  B  are  connected  by  a  system  of  pipes  as  shown  in 
Fig.  103.  Pipe  1  leading  from  reservoir  A  divides  at  S  into 
pipes  2  and  3  which  join  again  at  T.  Pipe  4  leads  from  the 
junction  T  to  a  reservoir  B.  Let  li,  d\  and  vi  be  respectively 
the  length,  diameter  and  mean  velocity  for  pipe  1,  and  the  same 
symbols  with  subscripts  2,  3  and  4  the  corresponding  quanitites 
for  pipes  2,  3  and  4.  $2  and  Qs  are  the  respective  discharges 
for  pipes  2  and  3,  the  sum  of  which  discharges  equals  Q,  the  total 
discharge  through  pipes  1  and  4.  Assuming  piezometer  tubes 


COMPOUND  PIPE  CONNECTING  TWO  RESERVOIRS      169 

at  £  and  T,  H  is  the  total  head  lost  in  the  system  of  pipes,  hi 
is  the  head  lost  in  pipe  1,  h2  =  hs  is  the  head  lost  in  pipes  2  or  3, 
and  h*  is  the  head  lost  in  pipe  4.  Before  taking  up  any  of  the 
special  problems  suggested  by  this  figure  a  general  analysis  will 
be  given. 

Any  of  the  formulas  for  determining  loss  of  head  due  to 
friction  may  be  employed.  It  will  be  found  advantageous  to 
use  a  formula  whose  coefficient  does  not  vary  with  the  diameter, 


FIG.  103. — Compound  pipe. 

but  a  fractional  exponent  for  v  will  be  objectionable.     Formula 
(20)  has,  therefore,  been  selected.     Since  h2  =  ^3,  from  formula  (20), 


12 


k 


(40) 


also 


=  ^  =  ^|=       and       ,3=^  = 
Ci2       TTC12  $3 


writing  these  values  of  v2  and  v%  in  equation  (40), 


k    16Q22  1 


h    16<%    •     •     •     (4D 


K2 


Since  in  formula  (20)  K  varies  only  with  the  velocity  and  not 
with  the  diameter,  the  error  introduced  by  assuming  K2  =  K% 
will  not  be  important  unless  v\  and  v2  are  widely  different.  Assum- 
ing them  equal  and  canceling, 


or 


(42) 


(43) 


170  FLOW  OF   WATER  THROUGH  PIPES 

Placing 

-     ......  <«> 

Q-3  =  FQ2,    ........     (45) 

and  since 

Q,     .......    (46) 


Q2+FQ2  =  Q,       ......     (47) 

and 


Therefore  to  determine  approximately  (within  the  limits  of  the 
error  introduced  in  assuming  K  to  be  a  constant)  the  quantity  of 
water  passing  through  pipe  2  divide  the  total  discharge  by  1+F 
and  the  total  discharge  minus  Q2  gives  Q3. 

The  expression  for  loss  of  head  may  be  written, 

H  =  h1+h2+h4  ...........     (49) 

The  losses  of  head  in  the  pipes  1,  2  and  4  may  be  expressed  by 
either  formula  (16)  or  (20).  Using  (20),  the  expression  for  lost 
head  becomes 

V22  U       V42 

~          '     (50) 


But 

V2=Q2=*Q2 

a2     ird22 
and 

Q     4Q 


--- 


............ 

Substituting  these  values  of  v\,  v2  and  ^i  in  equation  (50)  and 
reducing 

ICQ"/        Z,  „          fe                    Z*  \ 

ff  =  2^?  V     d?38  (H-^)A"»        di^j  '  ( 

If  formula  (16)  in  place  of  (20)  had  been  used  in  writing  equation 
(50)  the  above  formula  would  be 


COMPOUND  PIPE  CONNECTING  TWO  RESERVOIRS      171 

In  formulas  (54)  and  (55)  the  coefficients  are  not  constants. 
Ki,  K2,  and  K±  vary  with  the  velocity  and  fi,  /2,  and  /4  vary 
both  with  the  velocity  and  diameter.  Formula  (54)  will  be  more 
convenient  in  solving  problems  where  all  diameters  are  not  known 
since  KI,  K^  and  K±  do  not  depend  upon  the  diameter  for  their 
value.  Three  types  of  problems  are  explained  below. 

Case  1. — Having  given  the  lengths  and  diameters  of  all  pipes 
and  the  total  lost  head;  to  determine  Q. 

Determine  F  by  formula  (44),  then  determine  an  approxi- 
mate value  of  Q  from  formula  (54)  or  (55)  estimating  the  velocities 
in  pipes  1,  2  and  4  for  obtaining  trial  values  of  KI,  K^  and  K±, 
or /i,  /2  and  /4.  If  the  estimated  velocities  were  not  too  much  in 
error  this  solution  may  give  the  value  of  Q  as  accurate  as  is 
desired;  otherwise  with  the  value  of  Q  obtained  by  the  first 
solution  determine  the  velocities  and  corresponding  values  of 
coefficients  in  the  three  pipes  and  again  solve  equation  (54)  or  (55) 
for  Q.  The  second  solution  should  always  give  Q  within  the  desired 
degree  of  accuracy. 

To  check  the  result,  from  the  computed  value  of  Q  determine 
hi  and  7i4,  then  using  H—(hi-\-h4)  as  the  friction  loss  in  pipes  2 
and  3  compute  62  and  63-  The  results  obtained  should  show 
Q  approximately  equal  to  62+63.  If  closer  agreement  between 
the  computed  values  of  Q  and  62+63  is  required  than  may  be 
obtained  readily  by  this  method  the  results  may  be  adjusted  by 
trial  solutions.  Since  there  is  always  uncertainty  as  to  the 
proper  value  of  coefficients  to  use,  it  is  not  usually  desirable  to 
work  for  too  close  an  agreement. 

If  preferred  this  problem  may  be  solved  entirely  by  trial 
but  it  will  save  time  in  trial  solutions  to  determine  first  by 
formulas  (44)  and  (48)  the  portion  of  the  total  flow  that  passes 
through  one  of  the  branching  pipes.  Then  successive  values  of 
6  may  be  assumed  and  the  lost  head  in  each  pipe  computed  until 
the  sum  of  the  losses  in  the  three  pipes  equals  the  total  lost  head. 
A  final  check  should  be  made  to  see  that  6  equals  approximately 
62+63. 

Case  2. — Having  given  the  discharge,  diameters  and  lengths 
of  all  pipes;  to  determine  the  total  lost  head. 

The  lost  head,  H,  may  be  determined  from  equations  (44) 
and  (54). 

If  preferred  62  may  be  first  obtained  from  formulas  (44)  and 


172  FLOW  OF  WATER  THROUGH  PIPES 

(48)  and  Q%  from  formula  (45).  Then  using  these  discharges, 
compute  the  head  lost  in  pipes  2  and  3.  These  should  be  equal. 
If  the  computations  do  not  show  them  equal  adjust  the  dis- 
charges by  trial  (reducing  the  discharge  in  the  pipe  showing  the 
greater  loss  of  head  and  increasing  by  the  same  amount  the 
discharge  in  the  other  pipe) ,  and  again  compute  the  head  lost  in 
each  pipe.  Repeat  the  assumption  and  computations  until  the 
losses  of  head  in  each  of  the  pipes  become  the  same  or  until  the 
agreement  is  close  enough  for  the  purpose.  This  loss  of  head 
plus  the  loss  of  head  in  pipes  1  and  4,  which  may  be  computed 
in  the  usual  manner,  gives,  H,  the  total  lost  head. 

Case  8. — Having  given  the  discharge,  the  total  lost  head,  the 
lengths  of  all  pipes  and  the  diameters  of  three  pipes;  to  deter- 
mine the  other  diameter. 

Assume  that  the  diameter  of  pipe  2  is  to  be  determined. 
Compute  the  head  lost  in  pipes  1  and  4  by  one  of  the  formulas 
for  determining  loss  of  head  due  to  friction.  Deduct  from  the 
total  lost  head  the  sum  of  these  computed  losses.  With  this 
difference,  which  is  the  head  lost  in  each  of  pipes  2  and  3,  deter- 
mine Qs.  Then,  Qz  =  Q  —  Q%.  With  Q2  known,  and  the  lost 
head  determined  as  described  above,  compute  the  diameter  of 
pipe  2. 

If  the  diameter  of  one  of  the  single  pipes,  as  for  example, 
pipe  4,  is  to  be  determined,  compute  the  head  lost  in  pipe  1, 
as  described  in  the  preceding  paragraph  and  also  the  head  lost 
in  the  branching  pipes  2  and  3  as  described  under  Case  2.  The 
difference  between  the  total  lost  head  and  the  sum  of  the  above 
losses  is  the  head  lost  in  pipe  4,  from  which  the  diameter  of  this 
pipe  may  be  computed. 

107.  Pipes  of  More  than  One  Diameter  Connected  in  Series. 
— Fig.  104  represents  a  pipe  of  three  diameters  with  lengths  Zi, 
£2  and  Is',  diameters  d\t  d2  and  cfa;  and  velocities  01,  V2  and  ^3. 
The  total  loss  of  head,  H,  assuming  formula  (19)  for  friction 
loss  is 


u^iT'    '     '     (56) 
also 

_xJ  2  xJ_2  .rJ-2 

....     (57) 


PROBLEMS 


1731 


substituting  the  values  of  vi,  V2  and  v%  obtained  from  (57)  in  equa- 
tion (56)  and  transposing  the  following  formula  is  obtained: 


16Q2, 


.     (58) 


From  this  formula  with  all  quantities  but  one  given,  the  unknown 
quantity  may  be  obtained.     If  the  velocity  is  not  known,  values 


of  Kit 
solution. 


FIG.  104.  —  Pipe  of  three  diameters. 

and  K$  must  be  assumed  and  corrected  from  a  trial 
A  second  solution  of  the  problem  may  then  be  made. 

PROBLEMS 


1.  A  new  cast-iron  pipe  1200  ft.  long  and  6  in.  in  diameter  carries  1.3  cu.  ft. 
per  second.     Determine  the  frictional  loss. 

2.  Determine  the  discharge  of  the  pipe  described  in  Problem  1  if  it  dis- 
charges under  a  head  of  80  ft. 

3.  What  diameter  of  new  cast-iron  pipe,  1  mile  long  is  required  to  dis- 
charge 4.5  cu.  ft.  per  second  under  a  head  of  50  ft.? 

4.  What  diameter  of  concrete  pipe  8000  ft.  long  is  required  to  discharge 
40  cu.  ft.  per  second  under  a  head  of  8  ft.? 

5.  Determine  the    loss  of  head  due  to  sudden  enlargement  if  a  pipe 
carrying  2.0  cu.  ft.  per  second  suddenly  changes  from  a  diameter  of  (a)  6  in. 
to  8  in.,  (&)  6  in.  to  12  in.,  and  (c)  6  in.  to  18  in.     Also  determine  the  difference 
in  pressure  resulting  from  these  changes. 

6.  Solve  Problem  5  if  the  direction  of  flow  is  reversed  in  each  case. 

7.  A  cast-iron  pipe  12  in.  in  diameter  and  100  ft.  long  having  a  sharp- 
cornered  entrance  draws  water  from  a  reservoir  and  discharges  into  the  air. 
What  is  the  difference  in  elevation  between  the  water  surface  in  the  reservoir 
and  the  discharge  end  of  the  pipe  if  the  rate  of  discharge  is  16.0  cu.  ft.  per 
second? 

8.  If  the  pipe  described  in  Problem  7  connects  two  reservoirs,  both  ends 
beinsc  sharp-cornered  and  submerged,  other  conditions  remaining  the  same, 


174  FLOW  OF  WATER  THROUGH  PIPES 

determine  the  difference  in  elevation  of  the  water  surfaces  in  the  two  reser- 
voirs. 

9.  A  wood-stave  pipe  having  a  diameter  of  48  in.  is  laid  on  a  down-grade 
of  6  ft.  per  mile.     Determine  the  difference  in  pressure  between  two  points 
1  mile  apart  if  the  discharge  is  45  cu.  ft.  per  second. 

10.  A  wood-stave  pipe,  500  ft.  long,  is  to  be  designed  to  carry  60  cu.  ft. 
per  second  across  a  ravine  so  as  to  connect  the  ends  of  an  open  flume.     If 
the  difference  in  elevation  between  the  water  surfaces  in  the  two  ends  of 
the  flume  is  to  be  5  ft.,  determine  the  necessary  diameter  of  pipe,  assuming 
abrupt  changes  in  section,  and  neglecting  the  effect  of  velocity  in  the  flume. 

11.  A  concrete-pipe  culvert  90  ft.  long  and  3  ft.  in  diameter  is  built 
through  a  road  embankment.     The  culvert  is  laid  on  a  grade  of  1  ft.  per 
100  ft.     Water  is  backed  up  to  a  depth  of  5  ft.  above  the  top  of  the  pipe 
at  the  entrance  and  at  the  outlet  the  top  of  the  pipe  is  submerged  to  a  depth 
of  2  ft.     Assume  a  sharp-cornered  entrance.     What  is  the  discharge? 

12.  In  Problem  11  (assuming  all  other  conditions  to  be  the  same),  what 
diameter  of  pipe  will  be  required  to  discharge  100  cu.  ft.  per  second? 

13.  In  Problem  11  (assuming  all  other  conditions  to  be  the  same)  what 
will  be  the  depth  of  water  above  the  top  of  the  pipe  at  its  entrance  when 
the  culvert  is  discharging  50  cu.  ft.  per  second? 

14.  A  pipe  line  is  to  be  laid  connecting  two  tangents  which  intersect  at 
an  angle  of  90°.     Between  two  points  on  these  tangents,  each  distant  100  ft. 
from  their  point  of  intersection,  will  the  total  loss  in  head  be  less  if  a  bend 
having  a  radius  of  6  ft.  or  one  having  a  radius  of  50  ft.  is  used? 

15.  Three  new  cast-iron  pipes  are  connected  in  series  as  shown  in  Fig.  104. 
The  first  has  a  diameter  of  12  in.  and  a  length  of  1200  ft.;   the  second  has  a 
diameter  of  24  in.  and  a  length  of  2000  ft.;    and  the  third  has  a  diameter 
of  18  in.  and  a  length  of  1500  ft.     If  the  discharge  is  8  cu.  ft.  per  second, 
determine  the  lost  head  neglecting  the  minor  losses. 

16.  If,  in  Problem  15,  the  total  lost  head  in  the  three  pipes  is  45  ft., 
neglecting  the  minor  losses,  determine  the  discharge. 

17.  If  the  three  pipes,  described  in  Problem  15,  have  lengths  of  500  ft. 
each,  the  entrance  and  all  changes  in  section  being  sharp-cornered,  deter- 
mine the  total  lost  head  when  the  discharge  is  8  cu.  ft.  per  second. 

18.  Referring  to  Fig.  101,  page  166,  if  pipes  1,  2  and  3  have  diameters  of 
24  in.,  12  in.  and  18  in.,  and  lengths  of  1200  ft.,  500  ft.  and  1000  ft.,  respect- 
ively, determine  the  discharge  through  pipe  1,  if  ##  =  12  ft.  and  Hc  =  2Q  ft. 
Neglect  minor  losses. 

19.  In  Problem  18,  determine  HC,  if  the  discharge  through  pipe  1  is 
20  cu.  ft.  per  second,  all  other  conditions  remaining  the  same. 

20.  In  Problem  18,  determine  the  diameter  of  pipe  3  if  the  discharge 
through  pipe  1  is  24  cu.  ft.  per  second,  all  other  conditions  remaining  constant. 

21.  Referring  to  Fig.  103,  page  169,  if  pipes  1,  2,  3  and  4  have  diameters 
of  36  in.,  18  in.,  24  in.  and  30  in.,  and  lengths  of  3000  ft,,  2000  ft.,  2400  ft. 
and  1500  ft.,  respectively,  determine  H,  if  the  discharge  through  the  system 
is  60  cu.  ft.  per  second.     Neglect  the  minor  losses. 

22.  In  Problem  21,  determine  the  discharge  if  H  is  12  ft.,  other  con- 
ditions remaining  the  same. 


PROBLEMS  175 

23.  In  Problem  21,  determine  the  necessary  diameter  of  pipe  3  if  H  is 
15  ft.,  other  conditions  remaining  the  same. 

24.  Three  smooth  rubber-lined  fire  hose,  each  200  ft.  long  and  1\  in. 
in  diameter  and  having  1  in.  nozzles,  are  connected  to  a  6-in.  fire   hydrant 
If,  for  the  nozzles  Cc  =  l  and  Cp  =  0.97,  determine  the  necessary  pressure  in 
the  hydrant  in  order  to  throw  streams  100  ft.  high,  the  nozzles  being  10  ft. 
above  the  hydrant. 

25.  Determine  the  height  of  streams  that  can  be  thrown  if  the  pressure 
in  the  hydrant  is  70  Ibs.  per  square  inch,  all  other  conditions  remaining  as 
stated  in  Problem  24. 


CHAPTER  X 
FLOW  OF  WATER  IN  OPEN  CHANNELS 

108.  Description  and  Definition. — An  open  channel  is  a  con- 
duit which  conveys  water  without  exerting  any  pressure,  above 
atmospheric  pressure,  other  than  that  due  to  the  actual  weight  of 
water  carried.     Water  therefore  ordinarily  flows  in  an  open  channel 
with  a  free  water  surface,  though  for  an  enclosed  conduit,  like  a 
sewer  flowing  full,  water  may  touch  the  top  surface  without  exert- 
ing pressure.     In  this  case  it  is  classed  as  an  open  channel.     Exam- 
ples of  open  channels  are  rivers,  canals,  flumes,  and  sewers  and 
aqueducts  when  carrying  water  not  under  pressure. 

Open  channels  have  various  forms  of  cross-section.  Artificial 
channels  are  commonly  of  rectangular,  trapezoidal,  or  circular 
cross-section,  while  natural  streams  have  irregular  channels. 
Though  friction  losses  in  open  channels  follow  the  same  general 
laws  as  in  pipes  and  pipe  formulas  could  be  adapted  to  them, 
special  friction  formulas  for  open  channels  are  usually  employed. 

109.  Wetted  Perimeter  and  Hydraulic  Radius. — The  wetted 

perimeter  of  any  conduit  is  the 
line  of  intersection  of  its  wetted 
surface  with  a  cross-sectional  plane. 

In  Fig.  105  the  wetted  perimeter  is 
FIG.   105.-Cross-section  of  trape-   the  length  of  the  broken  Hne  abcd 

zoidal  channel.  T  ,  ,    .,    n 

In  a  circular  conduit  flowing  part 

full,  as  a  sewer,  the  wetted  perimeter  is  the  arc  of  a  circular  seg- 
ment and  in  a  natural  stream,  Fig.  116,  it  is  the  irregular  line 
abode. 

The  hydraulic  radius  of  any  section  of  a  channel  is  its  area 
divided  by  the  wetted  perimeter.  All  open  channel  formulas 
express  the  velocity  as  a  function  of  the  hydraulic  radius. 

110.  Friction  and  Distribution  of  Velocities. — As  described 
under  pipes  (Art.  91)  there  is  friction  between  the  moving  water 
and  the  surfaces  of  any  conduit.     If  there  were  no  other  influences 

176 


FRICTION  AND  DISTRIBUTION  OF  VELOCITIES        177 

the  maximum  velocity  would  ordinarily  occur  at  places  most 
remote  from  the  surfaces  of  the  conduit  which  produce  friction, 
and  consequently  at  the  water  surface.  Owing  to  surface  tension, 
however,  there  is  a  resistance  to  flow  at  the  surface  of  the  water, 
and  the  maximum  velocity  occurs  at  some  distance  below  the  sur- 
face. Under  ideal  conditions,  where  there  are  no  disturbing  influ- 
ences of  any  kind,  the  distribution  of  velocities  in  a  regular  channel 
will  be  uniform  and  similar  on  either  side  of  the  center.  There 
are  always,  however,  some  irregularities  in  every  channel  sufficient 
to  prevent  a  uniform  distribution  of  velocities.  The  lines  of  equal 
velocity  plotted  from  a  large  number  of  velocity  measurements  for 
the  Sudbury  conduit  near  Boston,  Fig.  106,  shows  a  more  regular 
distribution  of  velocities  than  will  be  found  in  most  channels. 
The  distribution  of  velocities  in  a  river  of  irregular  cross-section,  as 


FIG.  106. — Distribution  of  velocities  in  Sudbury  conduit. 

determined  from  velocity  measurements  with  a  current  meter,  is 
shown  in  the  upper  portion  of  Fig.  107.  The  figures  show  the 
velocities  obtained  at  the  points,  where  measurements  were  made 
and  the  irregular  lines  are  interpolated  equal  velocity  lines. 

The  curves  in  the  lower  portion  of  Fig.  107  show  the  distribu- 
tion of  velocities  in  vertical  lines.  These  curves  are  called  vertical 
velocity  curves  and  the  velocities  from  which  they  are  plotted  are 
called  velocities  in  the  vertical.  The  following  properties  of  ver- 
tical velocity  curves  have  been  determined  from  the  measurement 
of  velocities  of  a  large  number  of  streams  and  a  study  of  the  curves 
plotted  from  them. 

(a)  Vertical  velocity  curves  have  approximately  the  form  of 
parabolas  with  horizontal  axes  passing  through  the  thread  of 
maximum  velocity.  In  general  the  maximum  velocity  occurs 
somewhere  between  the  water  surface  and  one-third  of  the  depth, 


178 


FLOW  OF  WATER  IN  OPEN   CHANNELS 


the  distance  from  the  surface  to  the  point  of  maximum  velocity 
being  at  a  greater  proportional  depth  for  greater  depths  of  water. 
For  shallow  streams  the  maximum  velocity  is  very  near  to  the 
surface  while  for  very  deep  streams  it  may  lie  at  about  one-third 

Distances  in  feet 
0        10       20      30      40       50       60      70       80       90      100      UO     120     130     140      150  , 


CURVES  OF  EQUAL  VELOCITY 


VERTICAL  VELOCITY  CURVES 

Note;    Figures  at  top  of  curves  indicate  measuring  points 
Figures  at  bottom  of  curves  indicate  mean  velocity  in  the  vertical 
20      30  40       50  60    70         80  90     100      UO       120     130  140       150 


1.12 


-1.91 


1.99 


1.87 


2.22 


2.27 


-/ 

1.58 


2.28- 


2.25 


2.11 


1.88 


Horizontal  divisions  represent  1  foot  per  second  velocity 

FIG.  107. — Velocities  in  natural  stream. 


of  the  depth.     A  strong  wind  blowing  either  upstream  or  down- 
stream will  affect  the  distribution  of  velocities  in  the  vertical. 

(6)  The  mean  velocity  in  the  vertical  is  ordinarily  found  at 
a  distance  below  the  surface  varying  from  0.55  to  0.65  of  the  depth. 
The  velocity  at  0.6  depth  is  usually  within  5  per  cent  of  the  mean 
velocity. 


FRICTION  AND  DISTRIBUTION  OF  VELOCITIES        179 

(c)  The  mean  of  the  velocities  at  0.2  and  0.8  depth  usually 
gives  the  mean  velocity  in  the  vertical  within  2  per  cent. 

(d)  The  mean  velocity  in  the  vertical  is  ordinarily  from  0.80 
to  0.95  of  the  surface  velocity.     The  smaller  percentage  applies 
to  the  shallower  streams. 

The  above  properties  of  vertical  velocity  curves  are  made 
use  of  in  measuring  the  discharges  of  streams.  Mean  velocities 
in  successive  verticals  are  first  obtained  by  measuring  the  velocity 
at  0.6  of  the  depth  in  each  vertical  or,  where  greater  accuracy  is 
required,  by  taking  the  mean  of  the  velocities  at  0.2  and  0.8  of 
the  depth.  The  mean  of  the  velocities  in  any  two  adjacent  ver- 
ticals is  considered  as  the  mean  velocity  of  the  water  between  these 
verticals.  The  area  between  the  verticals  having  been  determined, 
the  discharge  through  this  portion  of  the  cross-section  of  the  stream 
is  the  product  of  this  area  and  the  mean  velocity.  The  sum  of  all 
discharges  between  successive  verticals  is  the  total  discharge. 

The  mean  velocity  in  the  vertical  may  be  obtained  by  taking 
the  mean  of  several  velocity  measurements.  This  method  is  more 
laborious,  however,  and  does  not  give  the  mean  velocity  appreciably 
more  accurately  than  that  obtained  by  taking  the  mean  of  velocities 
at  0.2  and  0.8  of  the  depth. 

The  distribution  of  velocities  in  an  ice-covered  stream,  Fig.  108, 
is  modified  by  the  effect  of  friction  between  the  water  and  the  ice. 
The  amount  of  this  friction  exceeds  the  skin  friction  of  a  free  water 
surface  and  the  maximum  velocity  therefore  occurs  nearer  mid- 
depth.  The  mean  velocity  in  the  vertical  for  an  ice-covered 
stream  is  not  at  0.6  depth  but  the  mean  of  velocities  at  0.2  and  0.8 
depth  gives  approximately  the  mean  velocity  the  same  as  for  a 
stream  with  a  free-water  surface. 

111.  Energy  Contained  in  Water  in  an  Open  Channel.  —  This 
subject  has  been  discussed  (Art.  72)  in  connection  with  the  velocity 
of  approach  for  weirs,  and  the  energy  of  water  in  a  pipe  is  discussed 
in  Art.  92.  In  open  channels  where  velocities  in  different  parts  of 
a  cross-section  are  not  the  same,  the  total  kinetic  energy  contained 
in  the  water  flowing  past  any  cross-section  is 


(1) 


W  being  the  total  weight  of  water  passing  the  cross-section  in  one 
second,  v  the  mean  velocity,  and  a  an  empirical  coefficient  depend- 


180 


FLOW  OF   WATER   IN   OPEN   CHANNELS 


ing  for  its  value  upon  the  distribution  of  velocities  in  the  channel 
but  always  greater  than  unity.  The  range  of  variation  in  a  has 
not  been  determined  but  in  channels  with  unobstructed  flow  it 
probably  lies  between  1.1  and  1.2.  This  matter  is  not  of  great 
importance  in  ordinary  hydraulic  problems.  Velocity  head  in 
open  channels  is  commonly  taken  as  the  head  due  to  the  mean 


Distances  in  feet 


CURVES  OF  EQUAL  VELOCITY 


0 

ft 
8  2 


s 

I4 

1   5 

!• 


VERTICAL  VELOCITY  CURVES 

Note:  Figures  at  top  of  curves  indicate  measuring  points 
Figures  at  bottom  of  curves  indicate  mean  velocity  in  the  vertical 


Horizontal  divisions  represent  1  foot  per  second  velocity 

FIG.  108. — Velocities  in  ice-covered  stream. 


velocity,  that  is,  a  is  assumed  to  equal  unity.  The  slight  error 
introduced  by  this  assumption  may  be  partially  eliminated  by  a 
proper  selection  of  coefficients. 

112.  Continuity  of  Flow  in  Open  Channels. — In  any  open 
channel  fed  by  a  constant  supply  of  water  there  is  the  same  rate  of 
flow  past  every  cross-section.  In  pipes  flowing  full,  Art.  93,  since 
water  is  practically  incompressible,  it  is  not  necessary  that  the 
supply  be  constant  in  order  to  have  the  same  rate  of  flow  past 


LOSS  OF  HEAD 


181 


every  cross-section  at  the  same  instant;  but  in  open  channels 
the  water  is  not  confined  and  a  variation  in  the  rate  of  supply  will 
cause  unequal  rates  of  flow  past  different  sections  of  the  channel. 
In  other  words,  in  order  to  have  continuity  of  flow  in  an  open 
channel  it  is  necessary  at  the  same  time  to  have  steady  flow,  but 
in  pipes  flowing  full  there  is  always  continuity  of  flow  regardless  of 
whether  the  flow  is  steady  or  variable. 

When  continuity  of  flow  exists  in  an  open  channel,  the  mean 
velocities  are  equal  at  all  cross-sections  having  equal  areas  but  if 
the  areas  are  not  equal  the  velocities  are  inversely  proportional  to 
the  areas  of  the  respective  cross-sections. 

Thus,  if  a\  and  vi,  and  «2  and  v»  are  respectively  areas  and  mean 


FIG.  109. — Open  channel  with  accelerating  velocity. 

velocities  at  any  two  cross-sections  in  an  open  channel  where  con- 
tinuity of  flow  exists, 


and 


113.  Loss  of  Head.  —  Figs.  109  and  110  represent  channels  of 
constant  cross-sections  receiving  water  at  uniform  rates  from 
reservoirs.  Potential  energy  changes  to  kinetic  energy  as  move- 
ment of  the  water  takes  place.  As  the  water  flows  down  these 
channels  a  portion  of  the  energy  is  lost  by  friction  between  the 
water  and  the  surfaces  of  the  channels.  If  there  were  no  friction 
^sses  of  any  kind  the  velocity,  vt  at  any  section,  s,  would  be 
vt=\/2gh,  in  which  h  is  the  vertical  distance  between  the  water 
surface  in  the  reservoir  and  the  water  surface  at  the  section  of  the 
channel.  The  velocity  of  the  water  would  thus  continue  to  accel- 
erate as  long  as  the  downward  slope  of  the  channel  continued. 


182  FLOW  OF  WATER  IN  OPEN  CHANNELS 

The  actual  conditions  of  flow  as  modified  by  friction  for  the 
case  illustrated  in  Fig.  109  are  as  follows.  The  water  receives  a 
certain  initial  velocity  at  a,  (Art.  85)  where  the  channel  leaves  the 
reservoir.  This  channel,  as  indicated  in  the  figure,  has  a  slope 
steeper  than  is  required  to  carry  the  water  with  the  initial  velocity 
which  it  receives  at  a.  The  velocity  therefore  accelerates  for  some 
distance,  but  a  portion  of  the  energy  which  the  water  contains  is 
used  in  overcoming  friction.  As  the  water  proceeds  down  the 
channel  at  a  continually  increasing  velocity,  a  point  is  finally 
reached  (since  frictional  resistance  increases  with  the  velocity,  as 
has  already  been  shown  for  pipes  and  will  be  shown  for  open 
channels)  beyond  which  the  energy  used  in  overcoming  friction  in 
any  reach  exactly  equals  the  potential  energy  contained  in  the 
water  within  the  reach  by  reason  of  the  slope  of  the  channel. 
After  this  point  has  been  reached,  approximately  at  b  in  the  figure, 


FIG.  110. — Open  channel  with  constant  velocity. 

the  water  will  flow  to  c  and  beyond  with  a  constant  velocity  as  long 
as  channel  conditions  remain  unchanged.  As  water  moves  from 
a  to  6  a  portion  of  its  potential  energy  is  continually  being  changed 
to  kinetic  energy  and  the  remainder  is  used  in  overcoming  friction. 
Between  6  and  c  the  kinetic  energy  remains  constant  and  all  of 
the  potential  energy  is  used  in  overcoming  friction. 

Fig.  110  illustrates  the  case  where  the  slope  is  no  greater  than  is 
required  to  carry  the  water  at  the  initial  velocity  which  it  receives 
at  a.  Under  these  circumstances,  as  long  as  the  channel  conditions 
remain  constant,  the  velocity  in  the  channel  will  be  the  same  at  all 
sections.  In  other  words  the  potential  energy  of  the  water  will  all 
be  used  in  overcoming  friction  and  the  kinetic  energy  will  remain 
constant.  This  case  is  the  one  most  commonly  encountered  in 
open  channel  problems. 

In  this  connection  it  is  important  to  bear  in  mind  that  the 
velocity  remains  constant  only  so  long  as  the  channel  conditions 


LOSS  OF  HEAD  183 

remain  constant.  Any  change  in  the  size  of  the  channel  changes  the 
velocity.  An  increase  or  decrease  in  the  slope  of  the  channel 
causes  a  corresponding  increase  or  decrease  in  velocity.  The 
velocity  is  also  modified  by  any  conditions  affecting  frictional 
resistance.  It  will  be  noted  that  velocity  conditions  for  open 
channels  are  different  than  for  pipes.  In  a  pipe  flowing  full,  under 
a  constant  head,  the  water  being  confined  and  considered  incom- 
pressible, the  mean  velocity  can  change  only  when  the  diameter  of 
the  pipe  changes.  In  an  open  channel,  however,  the  water  being 
unconfined,  the  velocity  changes  with  every  change  in  channel 
conditions. 

Losses  of  head  in  open  channels  are  in  every  respect  analogous 
to  losses  of  head  in  pipes.  In  addition  to  the  loss  of  head  due  to 
friction  between  the  moving  water  and  the  surface  of  the  channel 
there  is  a  loss  of  head  wherever  the  velocity  of  water  or  direction  of 
flow  is  changed.  The  same  symbols  will  be  used  to  represent 
losses  of  head  in  open  channels  that  are  used  to  designate  the  cor- 
responding losses  in  pipes  (Art.  94).  These  losses  for  open  channels 
are  as  follows: 

(a)  A  continuous  loss  of  head  throughout  the  channel  due  to 
friction  between  the  moving  water  and  the  surface  of  the  channel 
and  to  viscosity.  This  loss  is  commonly  referred  to  as  the  loss  of 
head  due  to  friction. 

(6)  A  loss  of  head  at  entrance  to  the  channel,  that  is,  where  a 
channel  takes  water  from  a  reservoir  or  other  body  of  compara- 
tively still  water. 

(c)  A  loss  of  head  at  discharge,  that  is,  where  a  channel  dis- 
charges into  a  reservoir  or  other  body  of  comparatively  still  water. 

(d)  A  loss  of  head  due  to  contraction  where  a  channel  changes 
to  a  smaller  cross-sectional  area  causing  an  increase  in  velocity. 
The  loss  of  head  at  entrance  to  a  channel  (referred  to  under  (6) 
above)  is  a  special  case  of  this  loss. 

(e)  A  loss  of  head  due  to  enlargement  where  a  channel  changes 
to  a  larger  cross-sectional  area  causing  a  decrease  in  velocity. 
The  loss  of  head  at  discharge  (referred  to  under  (c)  above)  is  a 
special  case  of  this  loss. 

(/)  A  loss  of  head  due  to  obstructions  of  any  kind  in  a  channel, 
such  as  gates,  bridge  piers  or  submerged  weirs. 

(g)  A  loss  of  head  due  to  curves  in  a  channel  in  addition  to  the 
loss  which  occurs  in  an  equal  length  of  straight  channel. 


184 


FLOW  OF  WATER  IN  OPEN   CHANNELS 


114.  Hydraulic  Gradient  or  Water  Surface. — The  hydraulic 
gradient  (Art.  95)  of  an  open  channel  coincides  with  the  water  sur- 
face. In  the  case  of  steady,  uniform  flow,  the  water  surface  is 
parallel  to  the  bottom  of  the  channel.  Any  changes  in  channel 
conditions  which  will  cause  either  an  increase  or  decrease  in 
velocity  will  cause  a  change  in  the  elevation  3f  the  water  surface 
the  same  as  a  change  in  velocity  in  a  pipe  will  cause  a  change  in 
the  hydraulic  gradient.  Changes  in  the  hydraulic  gradient 
of  a  pipe  line  resulting  from  changes  in  section  are  frequently 
of  minor  importance  and  need  be  considered  only  insofar  as 
they  affect  the  total  loss  of  head,  while  for  an  open  channel  the 
effect  of  any  changes  in  the  cross-section  should  be  thoroughly 
understood  and  designs  for  the  transition  of  the  water  from  one 
velocity  to  another,  should  be  worked  out  with  great  care. 


FIG.  111. 

Failure  to  provide  properly  for  changes  in  elevation  of  water  sur- 
face at  the  place  where  such  changes  occur  may  make  it  necessary 
for  the  channel  to  carry  water  at  a  depth  other  than  that  for 
which  it  was  designed  and  thus  interfere  with  its  satisfactory 
operation. 

115.  Loss  of  Head  Due  to  Friction  in  Open  Channels. — Fig. 
Ill  represents  the  condition  of  steady,  uniform  flow  in  a  straight 
channel.  Since  all  of  the  head,  hf,  is  used  in  overcoming  friction  in 
the  distance,  Z,  this  lost  head  is  a  measure  of  the  resistance  to 
flow.  The  ratio  h//l  is  called  the  slope,  and  is  represented  by  the 
symbol  s.  Since  friction  losses  in  open  channels  and  pipes  are 
of  the  same  character  they  are  governed  by  the  same  laws.  To 
make  the  general  laws  as  stated  for  pipes  on  page  145  apply  to  open 
channels  it  is  necessary  only  to  substitute,  respectively,  the  words 
channel  and  hydraulic  radius  for  pipe  and  diameter.  It  is  evident, 
therefore,  that  the  base  formulas  for  pipes  apply  equally  to  open 
channels.  Formula  (14),  page  146,  is  in  the  form  generally  used 


THE  CHEZY  FORMULA  185 

for  open  channels.      For   convenience    of  reference    it   is   here 
repeated. 

v  =  K"'rvsz  ......     .     .     :     (2) 

The  further  consideration  of  friction  losses  in  open  channels 
must  be  purely  empirical.  Numerical  values  of  empirical  coef- 
ficients and  exponents  and,  if  necessary,  modifications  in  the 
form  of  the  base  formula  must  be  derived  from  experimental  data. 
A  few  of  the  more  commonly  used  open  channel  formulas  are 
given  in  the  following  pages. 

116.  The  Chezy  Formula.  —  This  formula  as  stated  in  the  pre- 
ceding chapter  (Art.  97)  is,  with  proper  modification,  applicable 
either  to  open  channels  or  to  pipes,  though  it  was  originally 
designed  for  open  channels.  The  formula  as  written  by  Chezy  is 

.......     (3) 


in  which  v  is  the  mean  velocity,  r  is  the  hydraulic  radius,  and  s  is 
the  slope  of  water  surface.  The  Chezy  formula  is  of  the  same  form 
as  formula  (2)  ;  y  and  z  each  being  equal  to  J  and  C  being  written 
for  K'".  The  value  of  the  coefficient  C  varies  with  the  character- 
istics of  the  channel.  In  the  form  given  it  is  not  therefore  readily 
adaptable  to  open  channels  but  with  modifications  it  forms  the 
basis  of  most  of  the  formulas  in  common  use. 

117.  The  Kutter  Formula.  —  An  elaborate  investigation  of  all 
available  records  of  measurements  of  flow  in  open  channels  was 
made  by  Ganguillet  and  Kutter,1  Swiss  engineers,  in  1869.  As  a 
result  of  their  study  they  deduced  the  following  empirical  formula, 
commonly  called  the  Kutter  formula,  for  determining  the  value  of 
C  in  the  Chezy  formula. 

000281 


^ 

(A) 


Vr 

In  the  above  formula,  C  is  expressed  as  a  function  of  the  hydraulic 
radius,  r,  and  slope,  s,  as  well  as  the  coefficient  of  roughness,  n,  whose 
value  increases  with  the  degree  of  roughness  of  the  channel. 

1  GANGUILLET  and  KUTTER:  Flow  of  Water  in  Rivers  and  Other  Channels. 
Translation  by  HERRING  and  TRAUTWINE,  John  Wiley  &  Sons,  Publishers. 


186  FLOW  OF  WATER  IN  OPEN   CHANNELS 

VALUES  OF  C  FROM  KUTTER'S  FORMULA 


HYDRAULIC  RADIUS  r  IN  FEET 

Slope 

n 

0.2 

0.3 

0.4 

0.6 

0.8 

1.0 

1.5 

2.0 

2.5 

3.0 

4.0 

6.0 

8.0 

0.0 

5.0 

.00005 

0.010 

87 

98 

109 

123 

133 

140 

154 

164 

172 

177 

187 

199 

207 

213 

220 

.012 

68 

78 

88 

98 

107 

113 

126 

135 

142 

148 

157 

168 

176 

182 

189 

.015 

52 

58 

66 

76 

83 

89 

99 

107 

113 

118 

126 

138 

145 

150 

159 

.017 

43 

50 

57 

65 

72 

77 

86 

93 

98 

103 

112 

122 

129 

134 

142 

.020 

35 

41 

45 

53 

59 

64 

72 

80 

84 

88 

95 

105 

111 

116 

125 

.025 

26 

30 

35 

41 

45 

49 

57 

62 

66 

70 

78 

85 

92 

96 

104 

.030 

22 

25 

28 

33 

37 

40 

47 

51 

55 

58 

65 

74 

78 

83 

90 

.0001 

0.010 

98 

108 

118 

131 

140 

147 

158 

167 

173 

178 

186 

196 

202 

206 

212 

.012 

76 

86 

95 

105 

113 

119 

130 

138 

144 

148 

155 

165 

170 

174 

180 

.015 

57 

64 

72 

81 

88 

93 

103 

109 

114 

118 

125 

134 

140 

143 

150 

.017 

48 

55 

62 

70 

75 

80 

88 

95 

99 

104 

111 

118 

125 

128 

135 

.020 

38 

45 

50 

57 

.63 

67 

75 

81 

85 

88 

95 

102 

107 

111 

118 

.025 

28 

34 

38 

43 

48 

51 

59 

64 

67 

70 

77 

84 

89 

93 

98 

.030 

23 

27 

30 

35 

39 

42 

48 

52 

55 

59 

64 

72 

75 

80 

85 

.0002 

0.010 

105 

115 

125 

137 

145 

150 

162 

169 

174 

178 

185 

193 

198 

202 

206 

.012 

83 

92 

100 

110 

117 

123 

133 

139 

144 

148 

154 

162 

167 

170 

175 

.015 

61 

69 

76 

84 

91 

96 

105 

110 

114 

118 

124 

132 

137 

140 

145 

.017 

52 

59 

65 

73 

78 

83 

90 

97 

100 

104 

110 

117 

122 

125 

130 

.020 

42 

48 

53 

60 

65 

68 

76 

82 

85 

88 

94 

100 

105 

108 

113 

.025 

30 

35 

40 

45 

50 

54 

60 

65 

68 

70 

76 

83 

86 

90 

95 

.030 

25 

28 

32 

37 

40 

43 

49 

53 

56 

59 

63 

69 

74 

77 

82 

.0004 

0.010 

110 

121 

128 

140 

148 

153 

164 

171 

174 

178 

184 

192 

197 

198 

203 

.012 

87 

95 

103 

113 

120 

125 

134 

141 

145 

149 

153 

161 

165 

168 

172 

.015 

64 

73 

78 

87 

93 

98 

106 

112 

115 

118 

123 

130 

134 

137 

142 

.017 

54 

62 

68 

75 

80 

84 

92 

98 

101 

104 

110 

116 

120 

123 

128 

.020 

43 

50 

55 

61 

67 

70 

77 

83 

86 

88 

94 

99 

104 

106 

110 

.025 

32 

37 

42 

47 

51 

55 

60 

65 

68 

70 

75 

82 

85 

88 

92 

.030 

26 

30 

33 

38 

41 

44 

50 

54 

57 

59 

63 

68 

73 

75 

80 

.001 

0.010 

113 

124 

132 

143 

150 

155 

165 

172 

175 

178 

184 

190 

195 

197 

201 

.012 

88 

97 

105 

115 

121 

127 

135 

142 

145 

149 

154 

160 

164 

167 

171 

.015 

66 

75 

80 

88 

94 

98 

107 

112 

116 

119 

123 

130 

133 

135 

141 

.017 

55 

63 

68 

76 

81 

85 

92 

98 

102 

105 

110 

115 

119 

122 

127 

.020 

45 

51 

56 

62 

68 

71 

78 

84 

87 

89 

93 

98 

103 

105 

109 

.025 

33 

38 

43 

48 

52 

55 

61 

65 

68 

70 

75 

81 

84 

87 

91 

.030 

27 

30 

34 

38 

42 

45 

50 

54 

57 

59 

63 

68 

72 

74 

78 

.01 

0.010 

114 

125 

133 

143 

151 

156 

165 

172 

175 

178 

184 

190 

194 

196 

200 

.012 

89 

99 

106 

116 

122 

128 

136 

142 

145 

149 

154 

159 

163 

166 

170 

.015 

67 

76 

81 

89 

95 

99 

107 

113 

116 

119 

123 

129 

133 

135 

140 

.017 

56 

64 

69 

77 

82 

86 

93 

99 

103 

105 

109 

115 

118 

121 

126 

.020 

46 

52 

57 

63 

68 

72 

78 

84 

87 

89 

93 

98 

102 

105 

108 

.025 

34 

39 

44 

49 

52 

56 

62 

65 

68 

70 

75 

80 

83 

86 

90 

.030 

27 

31 

35 

39 

43 

45 

51 

55 

58 

59 

63 

67 

71 

73 

>  77 

THE  KUTTER  FORMULA 


187 


Some  of  the  values  of  n  as  published  by  the  authors  of  the  formula 
are  as  follows : 

n  =  0.010  for  well-planed  timber  or  neat  cement; 

n  =  0.012  for  common  boards; 

n  =  0.013  for  ashlar  or  neatly  joined  brickwork; 

n= 0.017  for  rubble  masonry; 

n  =  0.020  for  canals  in  firm  gravel; 

n  =  0.025  for  canals  and  rivers  in  good  condition; 

n  =  0.030  for  canals  and  rivers  with  stones  and  weeds; 

n  =  0.035  for  canals  and  rivers  in  bad  order. 

The  above  values  do  not  include  all  present  construction  mate- 
rials, and  later  experimental  data  have  shown  the  need  of  revising 

MORTON'S  VALUES  OF  THE  COEFFICIENT  OF  ROUGHNESS,  n,   FOR  KUTTER'S 
AND  MANNING'S  FORMULAS 


RANGE  o] 

p  VALUES 

Commonly 

Surface 

From 

To 

used 
values 

Vitrified  sewer  pipe       .        .  . 

0  010 

0  017 

0  013 

Common,  clay  drain  tile 

0  Oil 

0  017 

0  014 

Glazed  brickwork  

0  Oil 

0  015 

0  013 

Brick  in  cement  mortar  .      .  . 

0  012 

0  017 

0  015 

Neat-cement  surfaces 

0  010 

0  013 

Cement-mortar  surfaces  

0  Oil 

0  016 

0  015 

Concrete  pipe                 .  .             ... 

0  012 

0  016 

0  015 

Plank  flumes  planed 

0  010 

0  014 

0  012 

Plank  flumes,  unplaned  

0  Oil 

0  015 

0  013 

Plank  flumes  with  battens 

0  012 

0  016 

0  015 

Concrete-lined  channels 

0  012 

0  018 

0  015 

Rubble  masonry  

0.017 

0  030 

Dry  rubble      .  .                                        .    . 

0  025 

0  035 

Ashlar  masonry 

0  013 

0  017 

Smooth  metal  flumes  ....          

0  Oil 

0  015 

Corrugated  metal  flumes 

0  022 

0  030 

Earth  canals  in  good  condition  

0.017 

0  025 

0  0225 

Earth  canals  with  weeds  and  rocks  
Canals  excavated  in  rock 

0.025 
0  025 

0.040 
0  035 

0.035 
0  033 

Natural  streams  in  good  condition  

0.025 

0  033 

Natural  streams  with  weeds  and  rocks  .... 
gluggish  rivers  very  weedy  

0.035 
0.050 

0.060 
0  150 

188  FLOW  OF  WATER  IN  OPEN   CHANNELS 

them.  A  more  complete  list  of  values  of  "Kutter's  n" 
based  upon  later  data  and  practice,  and  showing  the  probable 
ranges  of  variation  has  been  prepared  by  Horton,  extracts  from 
which  are  tabulated  on  page  187. 

The  solution  of  the  Kutter  formula  may  be  obtained  from 
tables  which  usually  accompany  the  formula.  The  use  of  the 
Chezy  formula  with  the  Kutter  coefficient  thus  becomes  much 
simplified.  A  short  table  of  values  of  C  corresponding  to  different 
values  of  r,  s,  and  n  is  given  on  page  186.  Interpolations  are 
necessary  in  using  this  table. 

118.  The  Manning  Formula.  —  This  formula  was  first  suggested 
by  Manning  1  in  1890.  His  study  of  the  experimental  data  then 
available  led  to  the  conclusion  that  the  values  of  the  exponents 
y  and  z  (formula  (2))  which  best  represented  the  law  of  flow  in 
open  channels  were,  respectively  f  and  J.  Expressed  in  English 
units  the  Manning  formula  is 

1.486 

(5) 


This  may  be  considered  as  the  Chezy  formula  with 


The  coefficient  of  roughness,  n,  is  to  be  given  the  same  value  as  n 
in  the  Kutter  formula.  The  values  of  n  applicable  to  different 
channel  conditions  are  tabulated  on  page  187.  Expressed  in  metric 
units  the  Manning  formula  is 

(6) 


119.  Comparison  of  Manning  and  Kutter  Formulas.  —  Using 
the  same  value  of  n  in  each  case,  the  Kutter  and  Manning  formulas 
give  identical  results  for  r=l  meter  =  3.  28  feet.  This  may  be 
proved  by  substituting  3.28  for  r  in  each  formula.  It  will  be  found 
that  each  formula  then  reduces  to 


1  ROBERT  MANNING  :  Flow  of  Water  in  Open  Channels  and  Pipes.     Trans. 
lust.  Civ.  Eng.  of  Ireland,  1890,  vol.  20. 


COMPARISON  OF  MANNING  AND  KUTTER  FORMULAS     189 

Since  s  does  not  appear  in  this  equation,  it  follows  that  when  r 
equals  1  meter,  the  Kutter  formula  gives  the  same  value  of  C 
for  all  slopes. 

Further  investigation  shows  that  for  hydraulic  radii  less  than 
1  meter,  with  the  same  value  of  n  used  in  each  formula,  the  Kutter 
formula  gives  somewhat  higher  values  of  C  than  the  Manning 
formula.  For  hydraulic  radii  greater  than  1  meter  the  values  of  C 
obtained  by  the  Kutter  formula  are  in  some  cases  slightly  less  and 
in  others  slightly  greater  than  the  values  obtained  by  the  Manning 
formula. 

It  has  been  found,  however,  from  several  hundred  gagings  of 
open  channels  which  were  made  under  a  wide  range  of  conditions 
as  regards  shape,  size  and  variation  in  roughness  that  the  proper 
values  of  n  to  be  used  in  the  two  formulas  are  so  nearly  identical 
that  for  all  practical  purposes  the  same  values  may  be  used.  In 
other  words,  with  the  same  value  of  n,  problems  solved  by  means 
of  one  of  the  formulas  will  give  results  agreeing  very  closely  with 
those  obtained  by  using  the  other  formula. 

The  Kutter  formula  has  for  many  years  been  the  most  widely 
used  of  all  of  the  open-channel  formulas.  It  has  been  used  almost 
exclusively  in  the  United  States  and  England  and  more  commonly 
than  any  other  formula  in  other  parts  of  the  world.  The  Manning 
formula  has  been  used  for  a  number  of  years,  in  Egypt,  India,  and 
Australia  and  quite  recently  many  American  engineers  have  come 
to  see  its  advantages  over  the  more  cumbersome  Kutter  formula. 

It  is  because  of  the  established  use  of  the  Kutter  formula  and 
the  general  familiarity  of  engineers  with  the  type  of  channel 
represented  by  different  values  of  "  Kutter  's  n  "  that  there 
is  an  advantage  in  including  n  in  the  Manning  coefficient. 
Expressed  as  it  is,  engineers  familiar  with  the  Kutter  formula  may 
adopt  the  Manning  formula  without  the  necessity  of  familiarizing 
themselves  with  a  new  coefficient,  and  at  the  same  time  get  prac- 
tically the  same  results  as  with  the  formula  with  which  they  are 
familiar. 

Very  evidently  the  Manning  formula  could  be  written 


.......     (7) 

and  values  of  K  could  be  selected  for  different  types  of  channels 
the  same  as  values  of  n  are  now  selected.  It  is  to  be  hoped  that 
this  form  of  the  formula  will  eventually  come  into  general  use,  and 


190 


FLOW  OF  WATER  IN  OPEN   CHANNELS 


that  the  present  form  will  be  simply  a  step  in  the  transition  from 
the  use  of  the  Kutter  formula  to  the  use  of  the  simplified  form  of  the 
Manning  formula.  In  order  to  assist  in  determining  K  in  terms  of 
n  the  following  table  of  values  is  given : 


MANNING'S  K  IN  TERMS  OF  n.     K  = 


1.486 


n 

K 

n 

K 

n 

K 

n 

K 

n 

K 

n 

K 

0.009 

165 

0.016 

93 

0.023 

65 

0.030 

50 

0.044 

34 

0.070 

21 

.010 

149 

.017 

87 

.024 

62 

.032 

46 

.046 

32 

.075 

20 

.011 

135 

.018 

83 

.025 

59 

.034 

44 

.048 

31 

.080 

19 

.012 

124 

.019 

78 

.026 

57 

.036 

41 

.050 

30 

.085 

18 

.013 

114 

.020 

74 

.027 

55 

.038 

39 

.055 

27 

.090 

17 

.014 

106 

.021 

71 

.028 

53 

.040 

37 

.060 

25 

.095 

16 

.015 

99 

.022 

68 

.029 

51 

.042 

35 

.065 

23 

.100 

15 

The  Kutter  formula  shows  C  to  be  a  function  of  the  slope,  s, 
while  the  Manning  formula  does  not.  The  terms  involving  s  in  the 
Kutter  formula  were  introduced  to  make  the  formula  agree  with  the 
measurements  of  flow  of  the  Mississippi  river  by  Humphreys  and 
Abbott.  These  measurements  have  since  been  shown  to  be  in 
error  by  at  least  10  per  cent  and  to  this  extent  the  Kutter  formula 
is  based  upon  incorrect  data.  Later  experiments  do  not  verify 
the  conclusions  of  Ganguillet  and  Kutter  regarding  this  matter. 
The  value  of  C  in  the  Kutter  formula  is  not  materially  affected 
by  the  s  terms  unless  the  slope  is  very  small,  much  smaller  than  is 
ordinarily  used  in  designing  channels,  and  so  the  formula  has 
been  satisfactorily  used  for  the  conditions  ordinarily  encountered 
in  practice.  It  is  probable  that  the  Kutter  formula  would  have 
been  more  satisfactory  for  all  channels,  including  those  with  very 
small  slopes,  with  the  "  s "  terms  omitted.  That  this  is  so  is 
shown  quite  conclusively  by  some  recent  experiments  on  flow  in 
the  Chicago  drainage  canal.1 

The  foregoing  discussion  may  be  summed  up  briefly  with  the 
statement  that  the  Manning  formula  is  much  simpler  to  use  than 
the  Kutter  formula  and  that,  with  the  same  value  of  n,  it  gives 
practically  the  same  results  as  the  Kutter  formula  except  for  flat 

1  MURRAY  BLANCHARD  :  Hydraulics  of  the  Chicago  Sanitary  Districts 
Main  Channel.  Journal  of  the  Western  Society  of  Engineers,  Sept.,  1920. 


THE  BAZIN   FORMULA  191 

slopes.     In  the  latter  case  the  Manning  formula  appears  to  give 
more  accurate  results  than  the  Kutter  formula. 

120.  The  Bazin  Formula.  —  This  formula,  first  published  l  in 
1897,  like  the  Kutter  formula,  determines  the  value  of  C  in  the 
Chezy  formula.  It  considers  C  to  be  a  function  of  r  but  not  of  s. 
Expressed  in  English  units  the  formula  is 

157.6 
~  .......      8 


in  which  m  is  a  coefficient  of  roughness.     Values  of  m  given  by 
Bazin  are 

m  =  0.109  for  smooth  cement  or  planed  wood; 

m  =  0.290  for  planks,  ashlar  and  brick; 

m  =  0.833  for  rubble  masonry; 

ra=  1.540  for  earth  channels  of  very  regular  surface; 

m  =  2.360  for  ordinary  earth  channel; 

m  =  3.170  for  exceptionally  rough  channels. 

The  Bazin  formula  is  used  extensively  in  France  but  it  has  not 
been  generally  adopted  in  other  countries.  The  value  of  m  is 
subject  to  fully  as  wide  a  variation  as  n  in  the  Kutter  or  Manning 
formulas. 

121.  Open-channel    Formulas    in     General.  —  The    Kutter, 
Manning,  and  Bazin  formulas  are  the  best  known  and  most  widely 
used  of  the  open-channel  formulas.     There  are  a  large  number  of 
other  formulas  which  have  been  published,  •  and  many  of  these 
doubtless  possess  merit.     It  is  not  ordinarily  advisable,  however,  to 
use  any  except  the  commonly  accepted  formulas  unless  there  is 
very  good  reason  for  so  doing.     The  successful  use  of  any  open- 
channel  formula  requires  an  accurate  knowledge  of  conditions,  and 
judgment  in  the  selection  of  coefficients.     Even  the  most  expe- 
rienced engineers  may  expect  errors  of  at  least  10  per  cent  in  select- 
ing coefficients  with  corresponding  errors  in  their  results. 

122.  Detailed  Study  of  Hydraulic  Gradient  or  Water  Surface. 
—  In  the  investigation  of  minor  losses  in  pipes,  Art.  102,  it  has  been 
shown  that  these  losses  may  be  added  collectively  to  the  loss  of 
head  due  to  friction  and  that  this  sum,  which  represents  the  total 

1  Annales  des  Fonts  et  Chaussees,  1897. 


192  FLOW  OF  WATER  IN  OPEN   CHANNELS 

lost  head,  may  be  considered  as  a  unit.  The  losses  of  head  for 
open  channels  are  similar  to  those  for  pipes  with  the  exception 
that  they  must  be  provided  for  at  the  places  where  they  occur. 

Losses  of  head  in  open  channels  which  result  from  changes  in 
velocity  may  be  expressed  as  functions  of  the  velocity  head  the 
same  as  for  pipes.  Thus, 

V2  .     V2  V2 

ho  =  K0—,     hd  =  Kd—,     hc  =  Kc—,  etc. 

in  which  v  is  the  mean  velocity  in  the  channel  having  the  smaller 
cross-sectional  area.  While  these  losses  of  head  for  open  channels 
are  frequently  of  much  greater  importance  than  the  similar  losses  for 
pipes,  the  values  of  coefficients  for  determining  them  are  not  so 
well  established.  More  experimental  data  in  this  field  are  needed. 


FIG.  112. — Change  of  channel  to  smaller  section. 

Fig.  112  shows  the  change  in  water  surface  resulting  from  con- 
tracting the  cross-sectional  area  of  a  channel.  The  mean  velocity  in 
the  larger  channel  is  v\  and  in  the  smaller  channel  v.  It  is  assumed 
that  the  grades  of  the  two  channels  are  just  sufficient  to  maintain 
these  velocities;  that  is,  there  is  uniform  flow  in  each  channel. 

Kinetic  energy  is  always  obtained  at  the  expense  of  potential 
energy.  In  any  open  channel  the  drop  in  water  surface,  h  in  Fig.  112, 
occurring  at  any  change  in  section  measures  the  loss  in  potential 
energy  resulting  from  the  change.  A  portion  of  h,  hv  in  figure,  is 
used  in  producing  kinetic  energy,  that  is  in  increasing  the  velocity 
of  the  water.  The  remainder,  hc,  is  the  head  used  in  overcoming 
friction  losses  at  the  place  where  the  change  in  velocity  occurs. 
Referring  to  the  figure, 

*-!-!' '••<•' 


DETAILED  STUDY  OF  WATER  SURFACE  193 

in  which  v  and  Vi  are  the  mean  velocities  in  the  smaller  and  larger 
channels  respectively.  This  expression  is  also  obtained  by  writing 
Bernoulli's  equation  between  points  in  a  filament  on  either  side  of 
the  change  in  section  when  equal  velocities  at  all  points  in  a 
cross-section  are  assumed  for  each  channel. 
From  the  figure 

hc+-V,      .....     (10) 


and  in  a  manner  similar  to  that  employed  for  losses  due  to  contrac- 
tion in  pipes 

hc  =  Kc          ..........     (11) 


and  substituting  this  value  in  equation  (10) 

V2        V2       Vi2 


(       . 


There  are  no  experimental  data  from  which  Kc  for  open  channels 
can  be  determined  but  it  appears  reasonable  that  it  may  have  a 
value  corresponding  to  that  for  contractions  in  pipes.  The  max- 
imum value  for  a  sharp-cornered  entrance  may  thus  be  taken  as 
0.5  with  a  smaller  value  for  a  rounded  or  tapered  entrance.  With 
care  in  design  the  value  of  Kc  may  be  reduced  very  nearly  to  zero. 
Under  the  most  favorable  conditions  where  Kc  is  zero,  the  differ- 
ence in  elevation  of  water  surfaces  will  be  hv.  This  drop  in  water 
surface  should  always  be  provided  for  when  a  canal  changes  to 
a  smaller  section. 

Example.  —  Assume  entrance  conditions,  such  that  ^  =  0.25; 
the  velocity  in  the  upper  channel,  vi  =  2.0  ft.  per  second;  in  the 
lower  channel  z;  =  8.0  ft.  per  second.  Determine  the  drop  in  water 
surface.  From  equation  (12) 


If  a  canal  discharges  from  a  reservoir  or  other  body  of  com- 
paratively still  water  the  conditions  are  the  same  as  above  except 
that  vi  may  be  considered  zero.  In  the  above  problem  v\  could 
have  been  considered  zero  without  materially  affecting  the  result. 

Fig.  113  shows  the  change  in  water  surface  resulting  from  enlarg- 
ing the  cross-sectional  area  of  a  channel.  The  mean  velocity  in  the 


194  FLOW  OF  WATER  IN  OPEN  CHANNELS 

smaller  channel  is  v  and  in  the  larger  channel  vi.  It  is  assumed 
that  the  slopes  of  the  two  channels  are  just  sufficient  to  maintain 
these  velocities. 

The  velocity  v  being  greater  than  v\  there  is  a  loss  in  kinetic 
energy  with  a  resultant  gain  in  potential  energy.  If  there  were  no 
loss  of  energy  from  friction,  all  of  the  kinetic  energy  in  the  smaller 
channel  in  excess  of  the  kinetic  energy  in  the  larger  channel  would 
be  converted  into  potential  energy,  and  the  water  surface  in  the 
larger  channel  would  be  at  a  distance  h0  above  the  water  surface 
in  the  smaller  channel.  Since  some  energy  is  required  to  overcome 
friction  and  turbulence  losses  where  the  transition  in  velocities 
occurs,  the  actual  elevation  of  water  surface  in  the  larger  channel 
is  lower  than  it  would  be  if  there  were  no  friction  losses,  or,  as 
indicated  in  the  figure,  at  a  distance,  h,  above  the  water  surface 
in  the  smaller  channel.  The  vertical  distance  he  =  hv—h  thus  rep- 


FIG.  113.  —  Change  of  channel  to  larger  section. 

resents  the  loss  of  head  due  to  enlargement.    This  may  be  expressed 
algebraically  as  follows  : 


also 

h  =  h,-he  =  fg-V-^-he,      ....     (14) 

V2 

and  substituting  Kt  —  for  he  the  formula  may  be  written, 

v2          v2    vi2 


There  are  no  satisfactory  experimental  data  giving  values  of  Ke. 
In  general,  however,  it  is  known  that  for  abrupt  changes  in  velocity, 
very  little  of  the  kinetic  energy  in  the  smaller  channel  is  con- 
verted into  potential  energy.  In  other  words  nearly  all  of  this 
energy  is  lost  in  friction  and  turbulence  and  there  is  little  differ- 


DETAILED  STUDY  OF  WATER  SURFACE  195 

ence  in  elevation  in  the  water  surfaces  in  the  two  channels.  In 
this  case  Ke  approaches  very  near  to  unity.  By  exercising  care 
in  design  and  construction  and  making  velocity  changes  gradual 
so  as  to  produce  a  minimum  of  turbulence  the  value  of  Ke  may  be 
greatly  reduced. 

It  should  be  remembered  that  equations  (9)  to  (15)  are  approx- 
imations in  that  they  assume  the  kinetic  energy  in  each  channel  to 
be  that  due  to  the  mean  velocity,  that  is,  they  assume  equal 
velocities  at  all  points  in  a  cross-section  for  each  channel.-  This 
is  equivalent  to  assuming  a  in  equation  (1)  to  be  equal  to  unity. 
The  error  introduced,  thereby  is  however,  of  no  great  importance, 
especially  in  view  of  the  fact  that  a  correction  for  this  error  is 
necessarily  included  in  the  empirical  values  selected  for  Kc  or  Ke. 

The  most  common  types  of  obstructions  in  open  channels  are 
submerged  weirs,  gates,  and  bridge  piers.  Losses  of  head  resulting 
from  weirs  of  all  kinds  and  gates,  which  are  types  of  orifices,  are 
treated  in  earlier  chapters.  These  structures  are  commonly 
employed  to  deflect  water  from  main  canals  to  secondary  channels. 

Bridge  piers,  Fig.  114,  restrict  the  cross-sectional  area  of  a  chan- 
nel and  therefore  obstruct  the  flow.     The  loss  of  head,  hg,  or  what 
amounts  to  the  same  thing,  the  amount  that  the  water  will  be 
backed  up  by  piers  is  not 
sufficient  to  be  of  any  im- 
portance  except   for   com- 
paratively   high   velocities. 
The  most   important   case 
arises    in   determining    the 
backing-up  effect  of  bridge 
piers  during  flood  stages  of      FlG  114.— Bridge  pier  obstructing  flow, 
streams.     The  total  loss  of 

head  is  made  up  of  three  parts;  a  loss  of  head  due  to  contraction 
of  the  channel  at  the  upstream  end  of  the  piers,  a  loss  of  head 
due  to  enlargement  of  the  channel  at  the  downstream  end,  and 
an  increase  in  loss  of  head  due  to  friction  resulting  from  the 
increase  in  velocity  in  the  contracted  portion  of  the  channel. 
On  account  of  the  higher  velocity,  the  surface  of  the  water 
between  the  piers  is  depressed,  the  vertical  distance,  h,  measuring 
the  increase  in  velocity  head  plus  the  loss  of  head.  The  distance 
h  —  hg  is  a  measure  of  the  velocity  head  reconverted  into  static 
head. 


196 


FLOW  OF  WATER  IN  OPEN   CHANNELS 


The  quantitative  determination  of  losses  of  head  from  piers  is 
entirely  empirical.  Though  many  experimental  data  are  available, 
no  satisfactory  general  formula  for  the  solution  of  this  problem  is 
known.  In  general,  however,  piers  that  are  so  designed  as  to  allow 
the  changes  in  velocity  to  occur  gradually  with  a  minimum  amount 

of  turbulence,  cause  the  smallest  loss 
of  head.  Fig.  115  represents  two 
horizontal  sections  of  piers.  Sec- 
tion A  will  cause  less  turbulence  and 
consequently  less  loss  of  head  than 
section  B> 

Curves  or  bends  in  the  alignment  of 
a  channel  cause  a  loss  of  head.  For 
low  velocities  such  as  occur  in  earth 
canals  the  loss  in  head  is  slight  and 
ordinarily  no  allowance  is  made  for  it 
unless  the  curves  are  sharp  and  fre- 
quent. For  sharp  curves  in  concrete- 
lined  canals  or  flumes  designed  to 
carry  water  at  high  velocities,  an 
increase  in  the  slope  should  be  pro- 
vided. There  are  few  experiments  for 
determining  the  loss  of  head  in  curves,  but  the  data  for  bends  in 
pipes  (page  163)  may  be  used  as  a  guide.  Some  engineers  prefer 
to  correct  for  loss  of  head  at  curves  by  using  a  higher  coefficient  of 
roughness. 

123.  Hydraulics  of  Rivers. — Open-channel  formulas  do  not 
apply  accurately  to  natural  streams  since  the  channel  sections  and 
slope  of  water  surface  vary  and  the  flow  is  non-uniform.  At  the 
lower  stages,  streams  usually  contain  alternating  reaches  of  riffles 
and  slack  water.  During  high  stages  this  condition  largely  dis- 
appears and  the  water  surface  becomes  approximately  parallel  to 
the  average  slope  of  the  bottom  of  the  channel.  The  degree  of 
roughness  of  natural  streams  varies  greatly  within  short  reaches 
and  even  within  different  parts  of  the  same  cross-section.  This 
may  be  seen  from  Fig.  116,  which  illustrates  a  stream  in  flood  stage. 
The  channel  of  normal  flow,  abc,  will  probably  have  an  entirely 
different  coefficient  of  roughness  than  the  flood  plain  cde.  Also  the 
portion  of  the  left-hand  bank,  Fig.  116,  lying  above  ordinary 
high  water  may  be  covered  with  trees  or  other  vegetation  and  have  a 


FIG.  115. — Effect  of  shape  of 
bridge  piers  in  causing  tur- 
bulence. 


IRREGULAR  SECTIONS  197 

higher  coefficient  than  the  lower  portion.  Rocks  and  other  channel 
irregularities  are  frequent  and  cause  varying  conditions  of  turbu- 
lence, the  effect  of  which  on  the  coefficient  of  roughness  is  difficult 
to  estimate. 

There  are,  however,  times  when  the  engineer  must  estimate  as 
well  as  he  can,  the  carrying  capacity  of  a  natural  channel.  This 
is  done  by  making  a  survey  of  the  stream,  from  which  cross-sections 
may  be  plotted  and  the  slope  of  water  surface  may  be  determined. 
A  certain  reach  is  selected  for  which  are  obtained  an  average  cross- 
section  and  slope  of  water  surface,  the  computations  being  based 
upon  these  average  values. 

It  is  apparent  that  results  obtained  in  this  manner  will  be  very 
approximate,  and  that  the  degree  of  accuracy  obtained  will  depend 
largely  upon  the  ability  of  the  engineer  to  judge  the  effect  of  these 
varying  conditions  upon  the 
coefficient  of  roughness.  Bet- 
ter results  will  be  obtained 
for     natural     streams    that 
have  fairly  straight  and  uni- 
form channels   and  are  free    FIG.  116.— Natural  stream  in  flood  stage, 
from  conditions  causing  tur- 
bulence.    At  the  higher  stages  channel  irregularities  have  less 
effect  upon  the  slope  of  water  surface,  and  open-channel  formulas 
then  apply  more  accurately. 

124.  Irregular  Sections. — Open-channel  formulas  should  not 
be  applied  directly  to  sections  having  a  break  or  pronounced 
irregularity  in  the  wetted  perimeter.  Fig.  116,  which  illustrates 
a  stream  in  flood  stage  shows  a  break  in  the  wetted  perimeter 
at  c.  That  the  open-channel  formulas  do  not  apply  directly  to 
the  entire  cross-section  in  such  a  case  may  be  shown  by  the 
following  example.  The  Manning  formula  is  used,  though  any 
of  the  other  open-channel  formulas  will  show  substantially  the 
same  results.  Assume  s  =  0.00.1;  n  =  0.035;  the  length  of  the 
portion,  abc,  of  the  wetted  perimeter  =  200  ft.,  and  of  the  portion 
cde  =  300  ft. ;  the  area  of  the  portion,  abcm  of  the  cross-section  = 
3000  sq.  ft.  and  of  the  portion  mode  =  900  sq.  ft.  Then  for  the 

entire  cross-section,  a  =  3900,  p  =  500,  r  =  -  =  ~^^  =7.8,  and  by 

J)       oUU 

the  Manning  formula,  Q  =  20,600  cu.  ft.  per  second.  The  cross- 
section  may  now  be  divided  into  the  two  portions,  abcm  and 


198  FLOW  OF  WATER  IN  OPEN  CHANNELS 

mcde.     Assuming  the  depth  of  water  over  the  flood  plain  to  be 
3  ft.,  the  wetted  perimeter  of  the  portion  of  the  cross-section, 


a&cwwill  be  about  203  ft.     Then  r  =  -  =         =U.78  and  from 


the  Manning  formula,  Q  =  24,250  cu.  ft.  per  second.  This  indi- 
cates that  a  portion  of  the  channel  discharges  more  water  than 
all  of  it,  which  is  clearly  impossible. 

Where  an  open-channel  formula  must  be  applied  to  an  irregular 
section  such  as  that  indicated  in  Fig.  116,  it  is  necessary  to  divide 
the  cross-section  into  two  portions  and  compute  the  discharges 
for  each  portion  separately.  As  the  two  portions  of  the  channel 
will  differ  in  roughness,  different  coefficients  should  be  selected  for 
each. 

125.  Cross-section  of  Greatest  Efficiency.  —  The  most  efficient 
channel  cross-section,  from  a  hydraulic  standpoint,  is  the  one 
which,  with  a  given  slope  and  area,  will  have  the  maximum 
capacity.  This  cross-section  is  the  one  having  the  smallest  wetted 
perimeter,  since  frictional  resistance  increases  directly  with  the 
wetted  perimeter.  This  also  may  be  seen  from  an  examination 
of  one  of  the  open-channel  formulas.  Take  for  example  the 
equation  for  Q  as  given  by  the  Manning  formula 


Under  the  assumptions  made  a,  n,  and  s  are  constant.  Q  there- 
fore increases  with  r.  Since  r=—,  r  increases  as  p  decreases  and 

since  Q  varies  only  with  r  it  is  a  maximum  when  p  is  a  minimum. 
It  should  be  borne  in  mind  in  'this  connection  that  there  are 
usually  practical  objections  to  using  cross- 
sections  of  minimum  area  but  the  dimen- 
sions of  such  cross-sections  should  be  known 
and  adhered   to   as  closely  as  conditions 
appear  to  justify. 
FIG.  I17.-Semicircular         Of  a11  cross-sections,  having  a  given  area 
channel.  ^ne  semicircle,  Fig.  117,  has   the   smallest 

wetted  perimeter  and   it   is  therefore  the 

cross-section  of  highest  hydraulic  efficiency.  Semicircular  cross- 
sections  are  sometimes  used  for  concrete  or  brick  channels  but 
they  are  not  used  for  earth  channels. 


CROSS-SECTION  OF  GREATEST  EFFICIENCY 


199 


Trapezoidal  cross-sections,  Fig.  118,  are  used  more  commonly 
than  any  others.  They  are  the  only  practical  sections  for  earth 
canals,  and  masonry  and  wooden  conduits  are  usually  of  this  form. 
The  rectangular  section,  usually  used  for  wooden  flumes,  may  be 
considered  as  a  special  case  of  the  trapezoidal  section.  Properties 
of  trapezoidal  sections  and  methods  of  determining  sections  of 
greatest  efficiency  are  shown  in  the  following  analysis. 

e  b 

From  Fig.  118,     =  z  and      =  y,  °r  e  =  Dz  and  b  =  Dy.    Then 


and 
or 


p=(y+2Vl+z2)D,        (17) 

a  =  D2(z+y),  (18) 


D  = 


(19) 


(«)  (6) 

FIG.  118. — Trapezoidal  channels. 

Substituting  this  value  of  D  in  equation  (17), 
p=(y+2Vl+*).  '   a 


(20) 


Equating  the  first  derivative  with  respect  to  y  to  zero  and  reducing, 

y  =  2(VT+^-z), (21) 

b  =  2D(Vl+z2-z) (22) 


or 


From  which  may  be  obtained  the  relation  between  depth  of 
water  and  bottom  width  of  canal  of  the  most  efficient  cross- 
section  for  any  values  of  z. 

From  equations  (17)  and  (18), 


P          D(y+2Vl+z2)' 


(23) 


200  FLOW  OF  WATER  IN  OPEN  CHANNELS 

Substituting  y  from  equation  (21)  and  reducing 

r  =  D/2,        ..........     (24) 

or,  the  cross-section  of  greatest  efficiency  has  a  hydraulic  radius 
equal  to  one-half  the  depth  of  water. 

By  substituting  y  from  equation  (21)  in  equation  (20)  and 
reducing,  the  following  expression  is  obtained 


-z,       .....     (25) 
equating  the  first  derivative  with  respect  to  z  to  zero  and  reducing 

z  =  4=  =  tan30°- 

\/3 

It  may  be  seen  from  an  examination  of  equation  (22)  that 
when  z  =  tan  30°  the  length  of  each  side  is  equal  to  6,  Fig.  118  (c) 
and  the  section  becomes  a  half  hexagon.  Thus,  of  all  the  trape- 
zoidal sections  (including  the  rectangle),  for  a  given  area,  the  half 
hexagon  has  the  smallest  perimeter  and  it  is  therefore  the  most 
efficient  trapezoidal  cross-section. 

From  equation  (22)  are  obtained  the  following  relations  between 
bottom  width  and  depth  (for  different  side  slopes)  for  trapezoidal 
cross-sections  of  maximum  efficiency, 

e/D  =  z    0         i  i  1  1J  2  3  4 

b        2D  1.567)     1.247)    0.837)    0.617)    0.477)    0.327)     0.257) 

A  semicircle  having  its  center  in  the  middle  of  the  water  surface 
may  always  be  inscribed  within  a  cross-section  of  maximum 

efficiency.  This  is  illustrated  for  a 
trapezoidal  cross-section  in  Fig.  119. 
OA,  OB,  and  OC  are  drawn  from  a 
point  0  on  the  center  line  of  the 
water  surface  perpendicular  to  the 
sides  of  the  channel  EF,  FG,  and 
GH  respectively.  Let  EF  =  GH=x] 
FG  =  b  ;  OA  =  OC  =  R  ;  and  OB  =  D.  As  before  a  =  area  of  section 
and  p  =  wetted  perimeter.  Then  from  the  figure 


and 

p  =  2x+b. 


CIRCULAR  SECTIONS 


201 


And  since  (equation  (24))  the  hydraulic  radius  equals  one-half  the 
depth  of  water 

D 


From  which 


p~    2x+b 
R  =  D. 


That  is,  OA,  OB,  and  OC  are  all  equal  and  a  semicircle  with 
center  at  0  is  tangent  to  the  three  sides. 

126.  Circular  Sections.  —  The  maximum  discharge  from  a 
channel  of  circular  cross-section  occurs  at  a  little  less  than  full 
depth.  This  may  be  seen  from  an  examination  of  open-channel 
formulas.  The  discharge  by  the  Manning  formula  is 


Q= 


(16) 


a  being  the  cross-sectional  area.     In  the 
investigation  of  a  particular  channel  n  and      / 
s  will  be  constant.     From  Fig.  120,  R  being    / 
the  radius  of  the  circle, 


and 


sin  6. 


FIG.  120.-Circular 
channel. 


With  these  equations  an  expression  for  ar%  may  be  written,  dif- 
ferentiating which  and  equating  to  zero,  the  value  of  6  which 
makes  Q  a  maximum  is  found  to  be  57°  40'.  The  corresponding 
depth  of  water  is  D  =  0.938d.  Other  open-channel  formulas  will 
give  substantially  the  same  result.  This  means  that  a  pipe  carry- 
ing water  not  under  pressure,  when  free  from  obstructions  and 
laid  on  a  true  grade,  will  not  flow  full. 

127.  Non-uniform  Flow.— In  uniform  flow  the  velocity  past 
all  cross-sections  in  the  channel  is  constant.  This  condition  obtains 
in  ordinary  conduits  where  successive  cross-sections  are  uniform 
in  size  and  shape  and  the  slope  of  water  surface  is  parallel  to  the 
bed  of  the  channel.  There  are,  however,  certain  cases  Where  the 
velocities  are  being  accelerated  or  retarded,  that  is,  the  flow  is 
non-uniform,  although  the  same  quantity  of  water  passes  all 


202  FLOW  OF   WATER   IN   OPEN   CHANNELS 

cross-sections  so  that  continuity  of  flow  exists,  or  expressed  by 
symbols, 

,  etc. 


An  example  of  non-uniform  flow  is  illustrated  in  Fig.  121,  which  rep- 
resents a  canal  supplied  by  another  canal  or  reservoir.  Water 
enters  the  canal  at  a  certain  initial  velocity  which  may  be  computed 
by  the  method  described  in  Art.  85.  There  is  a  sudden  drop  in  the 
water  surface  at  the  entrance.  The  slope  of  the  canal  is  greater 
than  that  required  to  carry  the  water  at  its  initial  velocity  and 
the  velocity  therefore  continues  to  accelerate  until  it  becomes  equal 
to  the  velocity  at  which  the  channel  will  carry  water  under  condi- 
tions of  uniform  flow. 

Fig.  122  which  represents  a  canal  connecting  two  reservoirs  is 


FIG.  121. — Non-uniform  flow  in  channel  with  steep  slope. 

another  example  of  non-uniform  flow.  The  bottom  of  the  canal 
is  on  a  slope  different  from  that  which  the  water  surface  will 
attain.  The  total  head  producing  flow  is  H.  As  before,  there  is  a 
drop  in  the  water  surface  at  the  place  where  the  water  receives  its 
initial  velocity.  The  same  analysis  applies  to  the  two  cases  illus- 
trated in  Figs.  121  and  122. 

This  problem  may  best  be  analyzed  by  considering  the  channel 
to  be  divided  up  into  reaches  A,  B,  C,  etc.,  Figs.  121  and  122,  the 
computations  being  made  for  one  reach  at  a  time.  There  is  a 
certain  degree  of  approximation  introduced  in  doing  this  as  com- 
putations are  based  upon  an  average  cross-sectional  area,  but  by 
reducing  the  length  of  reach  considered  any  desired  degree  of 
accuracy  may  be  obtained.  The  following  nomenclature  is 
used: 


NON-UNIFORM  FLOW 


203 


1  = 


si 

s 

hi 


d\ 

60 

61 

VQ 

v\ 

v 

r 

z 

n 


length  of  reach  considered; 

slope  of  bottom  of  canal; 

average  slope  of  water  surface  in  reach; 

fall  of  water  surface  in  reach; 

loss  of  lead  due  to  friction  in  reach; 

depth  of  water  in  upper  end  of  reach; 

depth  of  water  in  lower  end  of  reach; 

bottom  width  of  trapezoidal  section  at  upper  end  of  reach; 

bottom  width  of  trapezoidal  section  at  lower  end  of  reach; 

mean  velocity  at  upper  end  of  reach  ; 

mean  velocity  at  lower  end  of  reach; 

mean  velocity  at  middle  of  reach; 

hydraulic  radius  of  section  at  middle  of  reach; 

slope  of  sides  of  canal  for  trapezoidal  section  ; 

coefficient  of  roughness  in  Manning's  formula. 


FIG.  122. — Non-uniform  flow  in  channel  with  flat  slope. 

From  Bernoulli's  equation,  assuming  equal  velocities  at  all  points 
in  a  cross-section. 

(26) 


From  Manning's  formula, 
1.486 


.    .    .    (27) 


and  approximately,  putting  V  = 


(28) 


Substituting  this  value  of  hf  in  equation  (26)  and  transposing, 

^Irl^  Vsit* (29) 

This  is  the  general  equation  for  non-uniform  flow. 


204  FLOW  OF  WATER  IN  OPEN   CHANNELS 

If  the  areas  and  wetted  perimeters  of  the  cross-sections  at  the 
upper  and  lower  ends  of  a  reach  and  also  the  drop  in  water  surface, 
hi,  are  measured,  the  velocity  at  one  end  of  the  reach  may  be 
expressed  in  terms  of  the  velocity  at  the  other  end,  as,  for  example, 

VQ  =  --  ,   and   the  other  velocity,   vi,   may   be   computed  from 
0*0 

equation  (29). 

Similarly  if  Q  and  the  cross-section  at  the  upper  end  of  a  reach 
are  known,  the  cross-section  at  the  lower  end  of  a  reach  may 
be  computed.  Assume  for  example  a  trapezoidal  cross-section. 
Then 

hi=do+8il-di     .....    ^    .     .     .     (30) 

zd0y    ......... 

.........    (32) 


and  for  the  average  section, 


In  the  right-hand  members  of  the  above  equations  61  and  di  are 
the  only  unknown  quantities,  and  one  of  these  must  be  assumed. 
The  known  quantities  and  the  assumed  value  of  61  or  d\  are  then 
substituted  in  these  equations  and  the  expressions  for  hi;  v0,  vi, 
and  r  thus  obtained  are  substituted  in  equation  (29)  which  equation 
may  be  solved  for  61  or  di  depending  upon  which  has  been  assumed. 
Suppose  for  example  that  di,  the  depth  of  water  at  the  lower  end 
of  the  reach,  has  been  assumed;  61  is  then  computed.  If  the  pro- 
portions of  canal  section  thus  obtained  are  not  satisfactory,  a  new 
value  of  di  may  be  assumed  and  61  may  be  recomputed.  Ordina- 
rily the  general  form  of  the  canal  will  be  well  enough  known  so 
that  recomputations  will  be  unnecessary. 

To  get  the  dimensions  of  other  cross-sections  the  above  process 
will  be  repeated,  the  section  at  the  lower  end  of  one  reach  becoming 
the  upper  cross-section  of  the  next  reach  below. 

128.  Backwater.  —  A  common  problem  in  non-uniform  flow 
occurs  where  water  is  backed  up  by  a  dam,  weir  or  other  obstruc- 
tion. Usually  it  is  required  to  determine  the  amount  that  the 


BACKWATER 


205 


water  surface  will  be  raised  at  certain  specified  distances  upstream 
from  the  obstruction.  Fig.  123  indicates  a  channel  whose  water 
surface  without  the  obstruction  would  be  the  line  mn;  with  the 
obstruction  the  water  surface  assumes  the  curved  line  abode.  This 
latter  line  is  called  the  backwater  curve.  It  is  required  to  determine 
the  position  of  sufficient  points  on  the  backwater  curve  so  that  it 
may  be  plotted. 

In  this  case  the  velocity  is  retarded  but  the  general  principles 
are  the  same  as  for  non-uniform  flow  with  accelerating  velocity, 
and  the  same  general  method  of  solution  may  be  followed.  The 
channel  is  divided  up  into  reaches  A,  B,  C,  etc.,  as  before,  but 
as  the  elevation  of  water  surface  at  the  obstruction  is  usually  given 
the  computations  are  begun  at  this  section  and  continued  upstream. 


FIG.  123.— Backwater. 

This  is  not  necessary,  however,  as  the  computations  may  be 
begun  at  any  section  with  a  known  elevation  of  water  surface  and 
be  carried  either  upstream  or  downstream.  The  nomenclature  is 
that  given  on  page  203. 

The  general  equation  for  non-uniform  flow  as  given  on  page  203 


,    ^vi  _vo 
2g     2g 


8.83r* 


(29) 


do,  the  depth  of  water  at  the  upper  end  of  the  reach  is  the  quantity 
sought.  With  this  determined  the  depth  at  the  upper  end  of  the 
next  reach  may  be  obtained  in  the  same  manner.  This  problem 
may  be  solved  by  equation  (29)  in  a  manner  practically  identical 
with  that  described  in  Art.  126,  the  only  difference  being  that  the 
computations  proceed  upstream  instead  of  downstream. 

One  of  the  commonest  applications  of  the  backwater  curve 


206  FLOW  OF  WATER  IN  OPEN  CHANNELS 

problem  is  to  the  determination  of  the  elevations  of  points  in  the 
backwater  curve  above  a  dam.  This  is  necessary  when  the  damage 
which  will  result  from  submerging  property  during  floods  or  the 
effect  of  backwater  on  some  power  plant  farther  upstream,  is  to  be 
determined.  Since  in  natural  streams  the  channels  are  irregular, 
average  sections  in  each  reach  and  also  average  velocities  are  used. 
It  is  usual,  therefore,  to  put  the  average  velocity  in  the  reach,  v, 
in  formula  (29)  in  the  place  of  v0  and  vi,  from  which  the  follow- 
ing simplified  expression  is  obtained  : 

In2v2 


This  is  simply  a  transposed  form  of  the  Manning  formula,  hi  in 
this  case  being  equal  to  hf. 

In  solving  backwater  problems  by  this  formula,  Q  and  the 
elevation  of  water  surface  at  the  lower  end  of  reach  A  are  usually 
known.  Then  hi  is  assumed,  from  which  a  trial  value  of  elevation 
of  water  surface  in  the  middle  of  the  reach  may  be  obtained.  Based 
upon  this  assumed  elevation  an  average  channel  cross-section  for  the 
reach  may  be  plotted  and  the  area  and  hydraulic  radius  determined. 
Then  hi  may  be  computed  and  if  this  computed  value  of  hi  differs 
from  the  assumed  value  sufficiently  to  materially  affect  the  results 
of  the  computations,  a  new  assumption  for  hi  may  be  made  and  the 
computations  repeated  until  the  assumed  value  of  hi  is  as  near  as 
is  desired  to  the  assumed  value.  With  a  little  experience  it  will  be 
found  that  the  first  assumed  value  of  hi  will  give  the  computed 
value  close  enough  without  repeating  the  computations. 

As  considerable  uncertainty  exists  in  the  selection  of  a  proper 
value  of  n  and  since  the  error  thus  introduced  into  the  result  is 
raised  to  the  second  power  any  solution  of  this  problem  is  neces- 
sarily approximate. 

129.  Divided  Flow.  —  Fig.  124  represents  a  channel  divided  by 
an  island.  The  total  discharge,  Q,  is  given  and  it  is  required  to 
determine  Qi  and  $2,  the  portion  of  discharge  going  respectively  to 
channels  1  and  2,  and  also  the  total  lost  head  hi}  that  is,  the  drop 
in  water  surface  from  m  to  n.  From  the  Manning  formula 


<»> 


DIVIDED   FLOW 


207 


and  since  v  =  —,  using  subscripts  1  and  2  to  refer  to  the  respective 
channels, 


hi 


and 


la     2.21r2K' 


(36) 


(37) 


FIG.  124.  —  Channel  divided  by  island. 


Equating  values  of  hi  and  reducing 


Putting 


and  since 


k 


(38) 


F  = 


nia2r2< 


Q2  =  T+F> 


(40) 
(41) 

(42) 


h\  may  now  be  obtained  by  substituting  Qi  and  Q%  in  formulas  (36) 
and  (37). 

In  computing  backwater  curves  where  the  channel  is  divided 
by  an  island  of  considerable  length  it  may  be  necessary  to  make 
computations  for  separate  reaches,  as  described  on  page  205.  In 
this  case,  Qi  and  Q2  may  be  determined  approximately  from 
formulas  (40)  and  (42)  using  an  average  cross-section  for  each 
channel  and  with  these  discharges  determined  the  slope,  or  rise  in 
water  surface,  hi,  may  be  computed.  If  the  values  Qi  and  Q2 
are  correct,  the  computations  should  show  hi  the  same  for  each 


208  FLOW  OF  WATER  IN  OPEN  CHANNELS 

channel.  If  the  computed  values  of  hi  do  not  agree,  the  com- 
putations should  be  repeated,  reducing  the  discharge  of  the  channel 
for  which  the  computations  give  the  greatest  value  of  hi  and  increas- 
ing the  discharge  of  the  other  channel  by  an  equal  amount.  The 
computations  are  repeated  until  they  give  approximately  the  same 
values  of  hi  for  both  channels. 

It  may  be  helpful,  in  order  to  reduce  the  number  of  trial 
solutions,  to  plot  values  of  Q  against  the  error  made  in  each  assump- 
tion. The  method  is  similar  to  that  for  divided  flow  in  pipes, 
described  on  page  167. 

The  problem  presented  by  channels  having  a  flood  plain, 
Fig.  116  is  similar  to  that  of  the  channel  divided  by  an  island. 
The  flow  over  the  plain  should  be  computed  separately  from  that 
for  the  main  channel,  as  discussed  on  page  207.  If  the  total  dis- 
charge is  known,  Qi  and  $2  being  respectively  the  portions  of 
the  flow  for  the  main  channel  and  flood  plain,  formulas  (40)  and 
(42)  may  be  used  to  determine  approximate  values  of  Qi  and  $2. 
With  these  approximate  discharges  determined,  the  two  values  of 
hi  may  be  computed,  using  the  proper  value  of  n  for  each  channel. 
If  these  values  of  hi  do  not  agree  the  computations  should  be 
repeated,  correcting  Qi  and  62  and  continuing  the  computations  in 
the  same  manner  as  that  described  above  for  channels  divided  by 
an  island. 

PROBLEMS 

1.  An  earth  canal  in  good  condition  having  a  bottom  width  of  12  ft. 
and  side  slopes  of  2  horizontal  to  1  vertical  is  designed  to  carry  180  cu.  ft. 
per  second  at  a  mean  velocity  of  2.25  ft.  per  second.     What  is  the  necessary 
grade  of  the  canal? 

2.  What  is  the  capacity  of  the  canal  in  Problem  1,  if  the  grade  is  2  ft. 
per  mile,  all  other  conditions  remaining  the  same? 

3.  Determine  the  depth  of  water  in  the  canal,  described  in  Problem  1, 
if  the  grade  is  2  ft.  per  mile,  other  conditions  remaining  as  stated. 

4.  Determine  the  bottom  width  of  the  canal  having  the  same  capacity 
and  side  slopes  as  the  canal  described  in  Problem  1,  that  will  give  the  most 
efficient  section. 

6.  A  circular  concrete  sewer  5  ft.  in  diameter  and  flowing  half  full  has  a 
grade  of  4  ft.  per  mile.  Determine  the  discharge. 

6.  In  Problem  5,  what  slope  in  feet  per  mile  must  the  sewer  have  if  the 
mean  velocity  is  to  be  8  ft.  per  second  when  flowing  full  capacity? 

7.  A  smooth-metal  flume  of  semicircular  cross-section,  has  a  diameter  of 
6  ft.  and  a  grade  of  0.005.     What  diameter  of  corrugated  metal  flume  will 
be  required  to  have  the  same  capacity. 


PROBLEMS  2|)9 

8.  An  earth  canal  carries  a  depth  of  water  of  6  ft.     The  canal  is  20  ft. 
wide  on  the  bottom  and  has  side  slopes  of  1.5  horizontal  to  1  vertical.     s  = 
0.0002.     Using  a  value  of  n  of  0.025  compute  the  discharge  by  the  Manning 
formula  and  with  this  discharge  determine  the  value  of  n  in  the  Kutter 
formula  and  also  the  value  of  m  in  the  Bazin  formula. 

9.  An  earth  canal  is  to  be  designed  to  carry  400  cu.  ft.  per  second  at  a 
mean  velocity  of  2.2  ft.  per  second.     The   sides  of  canal  have   a   slope  of 
2  horizontal  to  1  vertical.     The  depth  of  water  is  to  be  one-fourth  of  the  bot- 
tom width.     Assuming  that  the  canal  will  be  maintained  in  good  condition 
find  the  necessary  grade. 

10.  An  earth  canal  in  good  condition  carries  200  cu.  ft.  per  second  at  a 
velocity  of  2  ft.  per  second.     Side  slopes  of  canal  are  2  horizontal  to  1  vertical. 
The  depth  of  water  is  one-third  of  bottom  width  of  canal.     This  canal 
discharges  into  a  flume  with  a  tapered  entrance,  the  conditions  being  such 
that  the  loss  of  head  at  entrance  may  be  considered  to  be  one-half  of  what 
it  would  be  for  an  abrupt  change  in  section.     The  flume  is  7  ft.  wide  and 
has  vertical  sides.     The  slope  of  the  bottom  of  the  flume  is  such  that  it 
carries  a  depth  of  water  of  3.5  ft.     Determine  how  much  the  bottom  of  the 
flume  should  be  above  or  below  the  bottom  of  the  canal. 

11.  An  earth  canal  containing  weeds  and  grass  has  a  bottom  width  of 
15  ft.  and  side  slopes  of  2  horizontal  to  1  vertical.     The  depth  of  water  is 
4  ft.  and  the  slope  is  2.75  ft.  per  mile.     It  is  desired  to  change  the  section 
to  a  semicircular  concrete-lined  channel  having  a  slope  of  1.5  ft.  in  1000  ft. 
Determine  the  radius  of  the  semicircular  channel  if  it  flows  full.     If  the 
change  in  section  is  abrupt  and  sharp-cornered,  what  will  be  the  drop  in 
water  surface  where  the  change  in  section  occurs? 

12.  An  earth  canal  in  good  condition  has  a  bottom  width  of  10  ft.,  side 
slopes  of  1.75  horizontal  to  1  vertical,  a  grade  of  0.00025,  and  carries  140  cu.  ft. 
per  second.     A  gate  is  constructed  at  the  lower  end  of  the  canal  which  dis- 
charges freely  into  the  air.     The  coefficient  of  discharge  of  the  gate  is  0.83. 
The  gate  is  to  be  3.0  ft.  high.     How  wide  should  the  gate  be  to  maintain  a 
constant  depth  of  water  in  the  canal? 

13.  A   rectangular   flume   of   unplaned   timber   connects   two   reservoirs 
300  ft.  apart.     The  flume  is  16  ft.  wide  and  both  entrance  and  exit  are  sharp- 
cornered.     The  bottom  of  the  flume,  which  is  on  a  level  grade,  is  5  ft.  below 
the  water  surface  in  the  upper  reservoir  and  2  ft.  below  the  level  in  the  lower 
reservoir.     Determine  the  discharge. 

14.  A  rectangular  flume  of  unplaned  timber  carries  water  from  a  reser- 
voir.    The  width  of  flume  is  20  ft.,  the  length  is  1000  ft.  and  the  slope  is 
1  ft.  per  100  ft.     If  the  entrance  is  sharp-cornered  and  the  bottom  of  the 
flume  at  the  entrance  is  4  ft.  lower  than  the  water  surface  in  the  reservoir, 
determine  the  rate  of  discharge. 

15.  An  earth  canal  in  good  condition  is  60  ft.  wide  on  the  bottom  and 
has  side  slopes  of  2  horizontal  to  1  vertical.     One  side  slope  extends  to  an 
elevation  of  20  ft.  above  the  bottom  of  the  canal.     The  other  bank,  which 
is  a  practically  level  meadow  at  an  elevation  of  6  ft.  above  the  bottom  of  the 
canal,  extends  back  500  ft.  from  the  canal  and  then  rises  abruptly.     The 
meadow  is  covered  with  short  grass  and  weeds.     If  the  slope  of  the  canal 


210  FLOW  OF  WATER  IN  OPEN  CHANNELS 

is  2.2  ft.  per  mile,  determine  the  discharge  when  the  water  is  8  ft.  deep  in 
the  canal. 

16.  A  flume  built  of  planed  lumber,  with  vertical  sides,  8  ft.  wide  has  a 
grade  of  2  ft.  per  1000  ft.     A  sharp-crested  weir  3.5  ft.  high  is  constructed 
across  the  flume.     When  the  head  of  water  over  the  weir  is  3.1  ft.,  what 
is  the  elevation  of  water  surface  at  a  section  200  ft.  upstream  from  the  weir? 
How  much  higher  is  the  water  surface  at  this  section  than  it  would  be  with 
the  same  quantity  of  water  flowing  but  with  the  weir  removed? 

17.  A  canal  carries  300  cu.  ft.  per  second  of  water  at  a  depth  of  5.5  ft. 
The  water  in  this  canal  is  to  be  dropped  to  a  lower  elevation  through  a 
concrete  chute  of  rectangular  cross-section,  having  a  grade  of  1  ft.  in  10  ft. 
The  chute  is  to  carry  a  uniform  depth  of  water  of  3.0  ft.,  the  width  to  vary  as 
required  to  maintain  this  depth.     Determine  the  width  of  chute  at  entrance, 
at  100  ft.  below  the   entrance,  and  at  200  ft.  below  the    entrance.     Also 
determine  the  minimum  width  possible  for  this  depth  of  water. 

18.  A  canal,  58,000  ft.  long  (580  Sta.'s),  is   to  be  constructed  with  a 
capacity  of  300  cu.  ft.  per  second.     The  canal  diverts  from  a  river  and 
terminates  at  a  reservoir  into  which  it  discharges.     The  water  surface  in 
the  river  at  the  point  of  diversion  is  to  be  maintained  at  an  elevation  of 
770  ft. 

(a)  Water  is  to  be  diverted  through  six  head  gates,  having  rectangular 
orifices  each  2  ft.  by  5  ft.  Determine  the  head  required  to  force  the  water 
through  these  openings,  assuming  a  coefficient  of  discharge  of  0.80. 

(6)  From  Sta.  0  to  Sta.  425  the  canal  is  in  earth  section,  having  side 
slopes  of  2  horizontal  to  1  vertical,  and  a  depth  of  water  of  0.3  of  the  bottom 
width  of  the  canal.  Velocity  of  water  is  to  be  2.1  ft.  per  second.  Assume  a 
coefficient  of  roughness  of  0.0225.  Determine  grade  or  slope  of  canal. 

(c)  Between  Sta.  425  and  500  the  canal  is  in  rock  and  is  to  have  a  semi- 
circular section  lined  with  concrete.     The  grade  of  the  canal  is  to  be  2  ft.  per 
1000  ft.      Coefficient  of   roughness,   0.014.      Determine  the  head  lost   at 
entrance,  using  a  coefficient  of  velocity  of  0.92.     Also  determine  diameter  of 
canal  section. 

(d)  From  Sta.  500  to  Sta.  580  the  section  of  canal  is  the  same  as  from 
Sta.  0  to  Sta.  425.     At  the  reservoir  end  of  the  canal  (Sta.  580)  a  weir  is 
to  be  constructed  in  order  that  a  uniform  depth  of  water  may  be  maintained 
throughout  the  entire  length  of  earth  section.     Length  of  this  weir  is  to 
be  equal  to  the  bottom  width  of  the  earth  canal.     The  weir  has  a  rectangular 
section,  with  horizontal  crest  2  ft.  8  in.  wide.     Determine  height  of  crest 
above  bottom  of  the  canal. 

(e)  Tabulate  the  elevations  of  the  water  surface,  to  nearest  0.1  ft.,  at  the 
following  stations:    Sta.  0+10,  Sta.  424+90,  Sta.  425+10,  Sta.  500  and 
Sta.  579+90.     (Assume  Ke-l  at  Sta.  500.) 


CHAPTER  XI 
HYDRODYNAMICS 

130.  Fundamental    Principles. — Newton's    laws    of    motion 
form  the  basic  principles  of  the  subject  of  hydrodynamics.     These 
laws  are  clear  and  definite  and  lead  to  results  that  agree  exactly 
with  experiment.     Briefly  stated  they  are  as  follows: 

I.  Any  body  at  rest  or  in  motion  with  a  uniform  velocity  along  a 
straight  line  will  continue  in  that  same  condition  of  rest  or  motion 
until  acted  upon  by  some  external  force. 

II.  Any  change  in  the  momentum  of  a  moving  body  is  propor- 
tional to  the  force  producing  that  change  and  occurs  along  the 
same  straight  line  in  which  the  force  acts. 

III.  To  every  action  there  is  always  an  equal  and  opposite 
reaction. 

These  three  laws  of  Newton's  are  frequently  referred  to  as  the 
Laws  of  Inertia,  Force,  and  Stress,  respectively.  The  solution  of 
practically  any  problem  in  hydrodynamics  may  be  accomplished 
by  the  direct  application  of  these  laws.  It  is  therefore  essential 
that  a  clear  conception  be  had  of  their  full  significance.  As  an  aid 
in  acquiring  this  conception  the  following  discussion  is  presented. 

131.  Interpretation  of  Newton's  Laws. — Newton's  first  law  of 
motion  is  merely  a  statement  of  the  now  well-known  fact  that 
matter  is  inert;  that  is,  it  possesses  no  ability,  per  se,  to  change  its 
condition  of  rest,  or  motion,  and  that  any  such  change  must  be 
brought  about  through  the  action  of  some  external  force,  as  for 
instance,  friction  of  the  air  in  retarding  the  velocity  of  a  bullet. 

Since  change  in  motion  results  from  the  application  of  a  force 
it  may  be  assumed  that  the  magnitude  of  the  change  in  motion 
will  depend  upon  the  magnitude  of  the  force  producing  that 
change ;  in  other  words  it  may  be  assumed  that  there  is,  as  usual, 
a  direct  relation  between  cause  and  effect.  Making  this  assump- 
tion, the  second  law  follows  naturally  from  the  first.  Momentum 
is  by  definition  quantity  of  motion,  and  is  equal  to  the  product  of 

211 


212  HYDRODYNAMICS 

the  mass  and  velocity.     Designating  by  Z\,  the  momentum  of  a 
mass  of  water,  M,  having  a  velocity  v\. 

Zi=Mvi. 

If  a  force  P  acting   upon  this  mass  for  a  time  A£  changes  the 
velocity  to  v2. 

Z2  =  Mv2, 
and 

Z2-Z1=M(v2-v1),         (1) 

or 

&Z 
and 


since  —  =  acceleration,  and  mass  times  acceleration  equals  force. 
Equation  (2)  may  be  written  in  the  form, 

P=^l^        (3) 

or  substituting  for  Av  its  equivalent  (v2  —  vi)  and  letting  At  equal  1 
second,  equation  (3)  becomes, 

P  =  M(v2-vi) (4) 

This  means  that  when  a  force  acts  upon  a  mass  and  thereby 
changes  its  velocity  from  v\  to  v2,  the  force  is  equal  to  the  product 
of  the  mass  whose  velocity  is  changed  each  second  from  v\  to  v2, 
and  the  change  in  velocity. 

This  may  be  demonstrated  in  another  manner.  The  amount 
of  work  done  upon  any  mass  is  equal  to  the  gain  in  kinetic  energy, 
or 

Pl  =  —^ ^—  =-^-(v2  —  vi)(v2-\-vi),      ...     (5) 

Zi  Zi  Zi 

in  which  I  is  the  distance  through  which  the  force  P  acts  upon  the 
mass  M.  But 

"Y     ...  .     .     (6) 

Therefore,  from  equations  (1)  and  (2) 

),••..    .«•->...     (7) 


INTERPRETATION   OF  NEWTON'S  LAWS  213 

or  letting  t  =  1  second, 


(4) 


It  should  be  kept  clearly  in  mind  that  P  is  the  force  acting 
upon  the  mass  whose  velocity  is  changed  from  v\  to  V2.  Newton's 
third  law  of  motion  states  that  the  mass  reacts  with  a  force  equal 
in  magnitude  but  opposite  in  direction  to  the  force  P.  To  avoid 
confusion  the  reacting  force  that  water  exerts  upon  an  object  will 
be  designated  by  F.  Hence 

F=-P=-M(v2-vi)=M(vi-v2).        .     .     .     (8) 

Since  force  and  velocity  are  vector  quantities,  it  follows  that  if 
a  jet  of  water  impinges  against  a  vane  which  is  either  moving  or  at 
rest  and  thereby  has  its  velocity  in  any  direction  changed,  a  force  F 
is  exerted  upon  the  vane  whose  magnitude  in  any  direction  is  equal 
to  the  change  in  momentum  per  second  that  the  jet  undergoes 
in  the  same  direction.  In  other  words  the  force  F  is  equal  to  the 
mass  impinging  per  second  times  the  change  in  velocity  in  the 
direction  of  the  force.  The  X-  and  F-components  of  the  force 
exerted  by  a  jet  whose  path  lies  in  the  XY  plane  will  therefore 
be, 

^2  =  Mass  impinging  per  second  X  change  in  velocity  along  the 
X-axis. 

Fi,  =  Mass  impinging  per  second  X  change  in  velocity  along  the 
F-axis. 

F  =  \/FX2+FV2, 

and  the  tangent  of  the  angle  which  this  resultant  makes  with  the 

v        .     .     Fy 

A-axis  is  -=T. 

f  x 

The  change  in  velocity  may  be  either  positive  or  negative,  the 
only  difference  being  that  in  the  case  of  a  decrease  in  velocity  the 
dynamic  force  exerted  by  the  water  on  the  vane  is  in  the  same 
direction  as  flow,  whereas  in  the  case  of  an  increase  in  the  velocity 
the  dynamic  force  exerted  on  the  vane  is  opposed  to  the  direction 
of  flow.  For  instance,  referring  to  Fig.  127,  the  flow  being  from 
the  left  and  the  X-component  of  the  velocity  being  decreased,  Fx 
is  directed  toward  the  right,  whereas  the  F-component  of  the 
velocity  being  increased  and  the  flow  being  directed  upward,  Fv  is 
directed  downward. 


214 


HYDRODYNAMICS 


132.  Relative  and  Absolute  Velocities. — Strictly  speaking,  all 
motion  is  relative.  No  object  in  the  universe  is  known  to  be  fixed 
in  space.  An  airplane  is  said  to  be  flying  one  hundred  miles  per 
hour  but  this  is  its  velocity  only  with  respect  to  the  surface  of  the 
earth  beneath  it.  The  earth's  surface  itself  is  moving  at  a  tre- 
mendous speed  both  with  respect  to  its  axis  and  to  the  sun,  each 
of  which  are  whirling  through  space  at  a  still  greater  rate. 

It  is  nevertheless  convenient  in  connection  with  this  subject 
to  consider  all  motion  with  respect  to  the  earth's  surface  as  abso- 


TIG.  125. 

lute  motion.  The  airplane  above  referred  to  has  therefore  an  abso- 
lute velocity  of  100  miles  per  hour.  Another  plane  in  pursuit  may 
have  an  absolute  velocity  of  120  miles  per  hour  but  its  relative 
velocity  with  respect  to  the  first  plane  is  only  20  miles  per  hour. 
If  the  two  planes  were  to  fly  in  opposite  directions,  each  retaining 
its  same  absolute  velocity,  the  relative  velocity  between  them 
would  be  220  miles  per  hour.  If  they  were  to  fly  at  right  angles  to 

each  other  their  relative  ve- 
locities would  be  Vl002+1202 
=  156.2  miles  per  hour. 

Since  velocities  are  vector 
quantities,  these  results  may 
be  obtained  graphically  as  in 
Fig.  125,  in  which, 


VA  is  the  absolute  velocity 

of  A, 
VB  is  the  absolute  velocity 

of  B,  and 
u  is  the  relative  velocity  of 

either  with  respect  to 

the  other. 

133.  Jet  Impinging   Nor- 
mally on  a  Fixed  Flat  Plate.— 

Fig.  126  shows  a  jet  impinging  at  right  angles  against  a  fixed  flat 


FIG.  126. — Jet  impinging  against  a  flat 
plate. 


JET  IMPINGING  ON  A   MOVING  FLAT  PLATE  215 

plate.  It  is  assumed  that  the  plate  is  large  enough  with  respect  to 
the  size  of  the  jet  so  that  the  jet  is  deflected  through  a  full  90°. 
The  pressure  exerted  on  the  plate  varies  from  a  maximum  at  the 
axis  of  the  jet  where  it  is  very  nearly  equal  to  that  due  to  the  full 
velocity  head,  to  zero  at  a  distance  approximately  equal  to  the 
diameter  of  the  jet  from  its  axis.  The  total  pressure  exerted  is 
equal  to  the  product  of  the  mass  impinging  per  second  and  the  orig- 
inal velocity  of  the  jet,  since  the  final  velocity  of  the  water  as  it 
leaves  the  plate  has  no  component  in  its  original  direction. 
Hence,  the  force  exerted  on  the  plate  is, 

,  ,,.      wQ       wav2 
p  =  Mv  =  —  v  =  --  ,    ......  (9) 

Q  9 

where  M  and  Q  represent  respectively  the  mass  and  quantity 
striking  the  plate  per  second,  and  a  and  v  are  the  cross-sectional 
area  and  mean  velocity  of  the  jet,  all  terms  being  expressed  in 
the  foot-pound-second  system. 

134.  Jet  Impinging  Normally  on  a  Moving  Flat  Plate.  —  Con- 
sider the  case  of  a  jet  impinging  normally  against  a  flat  plate  mov- 
ing in  the  same  direction  as  the  jet  or  at  least  having  a  component 
of  its  motion  in  that  direction.  It  is  assumed  that  the  plate  has 
a  uniform  velocity,  being  restrained  from  accelerating  by  some 
external  agency.  The  mass  impinging  per  second  is 

.......     (10) 


where  Q'  is  the  quantity  of  water  striking  the  plate  in  cubic  feet 
per  second,  a  is  the  cross-sectional  area  of  the  jet  in  square  feet  and 
u  is  the  relative  velocity  of  the  jet  with  respect  to  the  plate.  The 
change  in  velocity  is  v—  v'  =  u,  since  the  velocity  in  the  direction 
of  the  force  is  changed  from  v  to  v'  and  the  jet  leaves  the  plate 
tangentially  with  a  velocity  whose  X-component  is  equal  to  the 
velocity  of  the  moving  plate.  The  force  acting  on  the  plate  is 

(11) 


9  9 

135.  Jet  Deflected  by  a  Fixed  Curved  Vane.  —  The  jet,  shown 
in  Fig.  127,  is  deflected  through  an  angle  6  by  a  fixed,  curved, 
trough-shaped  vane  A  B.  It  is  assumed  that  the  vane  is  so  smooth 
that  friction  may  be  neglected  so  that  the  velocity  with  which  the 


216 


HYDRODYNAMICS 


jet  leaves  at  B  may  be  considered  the  same  as  that  with  which  it 
strikes  at  A.  It  is  also  assumed  that  the  vertical  height  of  the 
vane  is  so  small  that  gravity  will  not  appreciably  retard  the  velocity 
of  the  jet.  Considering  the  horizontal  and  vertical  components 
of  the  force  acting  on  the  vane, 


=  ~—  (l-cos0), 
9 

sinfl. 


(12) 
(13) 


FIG.  127. — Jet  impinging  on  a  fixed 
curved  vane. 


FIG.  128. — Jet  impinging  on  a  curved, 
moving  vane. 


The  negative  sign  in  equation  (13)  means  that  the  force,  Fy, 
is  exerted  in  a  direction  opposite  to  vv,  or  in  other  words,  down- 
ward. 

The  resultant  force  is 


y 


sn 


(14) 


If  the  angle  of  deflection  is  greater  than  90°,  cos  0,  in  equation  (12) 
becomes  negative.  If  the  jet  is  deflected  through  a  full  180°, 
0=  —  1,  and  sin  0  =  0,  and  the  equations  become 

2wav2 


2Mv  = 


y 


(15) 
(16) 


136.  Jet  Deflected  by  a  Moving  Curved  Vane. — Consider  the 
vane  shown  in  Fig.  128  to  be  moving  with  a  uniform  velocity  v' 


JET  DEFLECTED  BY  A   MOVING  CURVED  VANE       217 


in  the  original  direction  of  the  jet.  The  absolute  velocity  of  the 
jet  as  it  impinges  at  A  is  v  and  its  relative  velocity  with  respect 
to  the  vane  is  v  —  v'  =  u.  The  mass  impinging  per  second  is  there- 
wau 


fore 


9 


Neglecting  friction  the  relative  velocity  of  the  jet  with 


respect  to  the  vane  remains  unchanged  while  flowing  from  A  to  B 
so  that  the  jet  leaves  the  vane  at  B  with  a  relative  velocity  u  in 
a  tangential  direction,  the  X-component  of  which  is  u  cos  6. 
The  change  in  velocity  along  the  X-axis  is  therefore  u— w  cos  0 
or  u(l  —  cos  B)  and 


(a) 


In  a  similar  manner  the  change  in  velocity  in  a  vertical  direc- 
tion is  from  zero  to  u  sin  B  and  therefore 


wau2 
9 


sin  6 


(18) 


The  absolute  velocity  and  direction  of  the  jet  as  it  leaves  the 
vane  are  shown  in  Fig.  128  by  the  vector  va  which  is  the  result- 
ant of  the  relative  velocity  u  and  the  velocity,  v',  of  the  vane. 

If  a  jet  is  directed  against  a  double-cusped  vane  as  shown  in 
Fig.  129  (a)  so  that  half  the  jet  is  deflected  by  each  cusp  through 
equal  angles,  Fx  will  be  determined  by  equation  (17)  but  Fy  is  zero 


218  HYDRODYNAMICS 

since  the  two  F-components  balance  being  equal  and  opposite  in 
direction. 

If  a  jet  is  deflected  through  a  full  180°  either  by  a  single-  or 
double-cusped  vane  as  shown  in  Fig.  129  (6)  and  (c),  obviously, 
for  a  stationary  vane, 


(19) 
y 
and  for  a  moving  vane 

(20) 


If  a  series  of  vanes  are  so  arranged  on  the  periphery  of  a  wheel 
that  the  entire  jet,  directed  tangentiallyjto  the  circumference  is 
striking  either  one  vane  or  another  successively,  the  mass  imping- 

ing again  becomes  M=  —  and  the  force  exerted  is, 
y 

F*  =  —  M(l-cos0),  ......     (21) 

l7 

Fv  =  —uswe  ........     (22) 

y 

It  should  be  noted  that  when  Fv  is  radial,  Fx  is  the  only  com- 
ponent of  the  force  tending  to  produce  rotation. 

137.  Work  Done  on  Moving  Vanes.  —  Since  work  is  equal  to 
force  times  distance  it  is  apparent  that  for  a  jet  to  do  any  work 
upon  a  vane,  the  vane  must  be  moving  with  a  velocity  between 
zero  and  the  velocity  of  the  jet  since  at  these  limiting  velocities 
either  the  distance  or  the  force  is  equal  to  zero.  The  question 
then  arises  as  to  what  velocity  the  vane  should  have,  for  any 
given  velocity  of  jet,  to  perform  the  maximum  amount  of  work. 

The  amount  of  work  done  per  second  is  the  product  of  the 
force  acting  in  the  direction  of  motion  and  the  distance  through 
which  it  acts.  Assuming  that  the  direction  of  motion  of  the  vane 
is  parallel  with  the  direction  of  the  jet,  the  force  acting  is  (Art.  136), 


(23) 


g 

and  the  distance  through  which  it  acts  per  second  is  equal  to  the 


WORK  DONE  ON   MOVING  VANES  219 

velocity  of  the  vane,  v'.     Representing  the  work  done  per  second, 
expressed  in  foot-pounds,  by  G, 


(24) 


Considering  v'  as  the  only  variable  in  this  expression  and  equat- 
ing the  first  derivative  to  zero,  the  relation  between  v  and  v'  may 
be  determined  for  which  G  is  a  maximum. 


from  which 


v'  =  v    and    z/  =   .....  (25) 

o 


When  v1  '  =  v,  no  work  is  done  since  the  force  exerted  is  then 
zero  and  this  value  represents  a  condition  of  minimum  work. 

For  maximum  work  therefore  v'  =  ^  . 

o 

In  the  case  of  a  series  of  vanes  so  arranged  that  the  entire  jet 
strikes  either  one  vane  or  another  successively,  the  force  exerted 
in  the  direction  of  motion,  which  is  assumed  parallel  with  the 
direction  of  the  jet  is  (Art.  136), 

Fx  =  —(v-v')(l-Cosd)  ......     (26) 

i/ 

The  distance  through  which  this  force  acts  in  one  second  is  v', 
and  therefore, 

G  =—(v-v')(l-cose)v'  .....     (27) 

J/ 


Differentiating,  and  equating  to  zero, 
dG    wav(l  —  cos  6) 


dv'  g 

and  for  maximum  work 
v 


(0-200=0, 


(28) 


220  HYDRODYNAMICS 

Substituting  this  value  of  vf  in  equation  (27), 

G-^d-oos*),      .......     (29) 

or 

G  =  ^(l-cos0),       .......     (30) 

which  is  -  -  ~  -  -  times  the  total  kinetic  energy  available  in  the 

2i 

jet.    For  0  =  180°  this  expression  equals  unity  and 


the  total  kinetic  energy  of  the  jet  being  converted  into  work. 
This  also  appears  from  considering  that  the  relative  velocity  of 

the  jet  as  it  leaves  the  vane  is  »,  which,  is  also  the  velocity  of  the 

vane.  These  two  velocities  being  equal  and  opposite  in  direction 
have  a  resultant  of  zero.  The  water  thus  leaves  the  vane  with 
zero  velocity,  signifying  that  all  of  its  original  energy  has  been 
utilized  in  performing  work. 

The  above  principles  are  made  use  of  in  the  design  of  impulse 
turbines,  which  consist  of  a  series  of  vanes  attached  to  the  per- 
iphery of  a  wheel.  The  angle  6  must  be  somewhat  less  than  180° 
so  that  the  jet  in  leaving  a  vane  will  not  interfere  with  the  suc- 
ceeding vane.  Making  the  angle  6  equal  to  170°  in  place  of  180° 
reduces  the  force  applied  to  the  wheel  by  less  than  1  per  cent. 

138.  Forces  Exerted  upon  Pipes.  —  In  the  preceding  articles 
of  this  chapter  the  discussion  has  been  restricted  to  forces  exerted 
by  jets  impinging  against  flat  and  curved  surfaces.     As  it  was 
always  considered  that  the  flow  was  free  and  unconfined  the  only 
forces  acting  were  dynamic. 

Consideration  will  now  be  given  to  the  longitudinal  thrust 
exerted  upon  a  section  of  pipe  by  water  flowing  through  it  under 
pressure.  This  thrust  will  usually  be  found  to  be  the  resultant  of 
both  static  and  dynamic  forces.  The  transverse  forces  which 
determine  the  necessary  thickness  of  pipe  were  discussed  in 
Art.  29. 

139.  Straight  Pipe  of  Varying  Diameter.  —  Under  conditions  of 
steady  flow  through  a  straight  pipe  of  varying  diameter  there  is  a 


STRAIGHT  PIPE  OF   VARYING   DIAMETER 


221 


longitudinal  thrust  exerted  upon  the  pipe.  This  thrust  is  the 
resultant  of  a  dynamic  force,  a  static  pressure,  and  frictional 
resistance. 

Fig.  130  shows  a  straight  section  of  converging  pipe.  Let  p\, 
ait  and  v\  represent  respectively  the  pressure,  area,  and  mean  veloc- 
ity at  A B  and  ^2,  «2,  and  V2  the  corresponding  values  at  CD. 
In  flowing  from  A  B  to  CD  the  water  is  accelerated  from  vi  to  V2 
and  the  force,  P,  producing  this  acceleration  is  the  resultant  of 
all  the  component  forces  acting  on  the  mass  A  BCD.  These  forces 
consist  of  the  pressures  on  the  sections  AB  and  CD,  the  pressure 
exerted  by  the  pipe  walls  ACBD  and  the  force  of  gravity,  the  last 


FIG.  130. 


of  which  can  be  neglected  since  it  acts  vertically  and  has  no  com- 
ponent in  the  direction  of  acceleration.  The  pressures  on  AB  and 
CD  are  a\,p\  and  a^pz  respectively,  a\p\  acting  in  the  direction 
of  acceleration  and  a^p^  being  opposed  to  it.  The  pressure,  dR, 
exerted  on  the  water  by  any  differential  area,  da,  of  the  pipe 
walls  will  be  inclined  slightly  from  the  normal  toward  the  direction 
of  flow  on  account  of  friction.  The  vertical  component  of  dR, 
being  normal  to  the  direction  of  acceleration  may  be  neglected, 
leaving  dRx  as  the  only  component  to  be  considered.  All  the 
values  of  dRx  for  the  various  elementary  areas  of  pipe  wall,  being 
parallel  and  acting  in  the  same  direction,  may  be  combined  into 
the  resultant,  Rx,  whose  magnitude  is  as  yet  unknown  but  whose 


222 


HYDRODYNAMICS 


direction  is  opposed  to  acceleration.     Therefore  the  force  pro- 
ducing acceleration, 

Rx  ........     (32) 


But  in  Art.  131  it  was  shown  that 


and  therefore,  from  equations  (32)  and  (33), 


y 


(33) 


(34) 


Since  Rx  is  the  X-component  of  the  forces  exerted  upon  the  water 
by  the  pipe  it  follows  that  the  thrust  exerted  upon  the  section  of 
pipe  by  the  water  must  be  equal  and  opposite  to  Rx,  or  in  other 
words  the  thrust  will  act  toward  the  right  in  Fig.  130. 

Considering  a  straight  section  of  pipe  of  constant  diameter 
throughout,  equation  (34)  reduces  to 


(35) 


since  0,1  =  0,2  =  a,  and  Vi  =  v2.     In  equation  (35)  pi  —  p2  is  the  drop 
in  pressure  resulting  from  friction  between  sections  A B  and  CD. 
140.  Pipe  Bends. — The  thrust  exerted  upon  a  curved  section 

of  pipe  of  either  constant  or 
varying  diameter  is  the  re- 
sultant of  component  forces 
similar  to  those  discussed  in 
the  preceding  article.  The 
chief  difference  lies  in  the  fact 
that  the  resultant  thrust  on 
a  curved  section  of  pipe  has 
both  X-  and  F-components 
since  there  is  a  change  in  ve- 
locity along  both  of  these  axes. 


Fig.     131    shows    a    pipe 
bend,  having  a  diameter  de- 
creasing from  AB  to  CD  and 
FlG  131  a  deflection  angle,  6.     Let  Rx 

and   Ry  represent  the  X-  and 
F-components  of  the  forces  exerted  by  the  pipe  upon  the  water. 


WATER  HAMMER  IN   PIPE   LINES  223 

The  X-  and  F-components  of  the  thrust  exerted  by  the  water  upon 
the  pipe  are  equal  in  magnitude  but  opposite  in  direction  to  Rz 
and  Ry  respectively.  Assuming  that  the  bend  lies  in  a  horizontal 
plane  so  that  the  action  of  gravity  is  normal  to  the  direction 
of  acceleration  and  therefore  may  be  ignored,  the  resultant  X  and 
Y  forces  producing  acceleration  are, 

—  C12P2  cos  6  —  Rx  =  —  (v2  cos  0  —  vi),       .     (36) 

Ry=—  v2  sin  0,    .....     (37) 
J 

the  right-hand  members  in  these  equations  representing  the  in- 
crease in  momentum  along  the  X-  and  F-axes  resulting  from  the 
accelerating  forces. 

From  the  above  equations, 

—  U2P2  cos  0-f—  (vi  —  v2  cos  0),     .     .     .     (38) 

t/ 

sin0+—  v2sme  ........     (39) 

\J 

If  the  pipe  bend  is  one  of  constant  diameter  throughout, 
ai=a2,  vi=V2t  and  pi=p2  (approximately),  and  the  equations 
reduce  to 

-cos*),      .......     (40) 

h*  ..........     (41) 


If  the  angle  0  equals  90°  these  equations  become 

(42) 


y 

141.  Water  Hammer  in  Pipe  Lines.  —  In  Fig.  132  is  shown  a 
pipe  line  leading  from  a  reservoir,  A,  and  discharging  into  the  air  at 
B  near  which  is  located  a  gate  valve.  If  the  valve  is  suddenly 
closed  a  dynamic  pressure  is  at  once  exerted  in  the  pipe  in  excess  of 
the  normal  static  pressure.  The  magnitude  of  this  pressure  is 
frequently  much  greater  than  that  of  any  static  pressure  to  which 
the  pipe  may  ever  be  subjected  and  the  possibility  of  the  occurrence 


224 


HYDRODYNAMICS 


of  such  pressure  must  therefore  be  investigated  in  connection  with 
the  design  of  any  pipe  line  of  importance. 

This  dynamic  pressure,  commonly  called  water  hammer,  is 
the  result  of  a  sudden  transformation  of  the  kinetic  energy  of  the 
moving  mass  of  water  within  the  pipe  into  pressure  energy.  Since 
force  equals  mass  times  acceleration,  or 


M- 
M      ' 


(43) 


it  follows  that  if  the  velocity  of  the  mass  M  could  be  reduced  from 
v  to  zero  instantaneously,  this  equation  would  become 


P  =  M 


0' 


(44) 


or  in  other  words  the  pressure  resulting  from  the  change  would  be 
infinite.     Such  an  instantaneous  change  is,  however,  impossible. 


FIG.  132. 


Consider  the  conditions  within  the  pipe  immediately  following 
the  closure  of  the  valve.  Let  l\,  1%,  Is,  ...  In  represent  infmitesi- 
mally  short  sections  of  pipe  as  shown  in  Fig.  132.  The  instant 
the  valve  is  closed,  the  water  in  contact  with  it  in  section  l\  is 
brought  to  rest,  its  kinetic  energy  is  transformed  into  pressure 
energy,  the  water  is  somewhat  compressed  and  the  pipe  wall 
with  which  it  is  in  contact  expands  slightly  as  a  result  of  the 
increased  stress  to  which  it  is  subjected.  Because  of  the  enlarged 
cross-sectional  area  of  l\  and  the  compressed  condition  of  the  water 
within  it,  a  greater  mass  of  water  is  now  contained  within  this  sec- 
tion than  before  the  closure.  It  is  evident  then  that  a  small  volume 
of  water  flowed  into  section  h  after  the  valve  was  closed.  An 


WATER   HAMMER   IN   PIPE   LINES  225 

instant  later  a  similar  procedure  takes  place  in  12  and  then  in  k,  so 
that  evidently  a  wave  of  increased  pressure  travels  up  the  pipe 
to  the  reservoir.  The  instant  this  wave  reaches  the  reservoir  the 
entire  pipe  is  expanded  and  the  water  within  it  is  compressed 
by  a  pressure  greater  than  that  due  to  the  normal  static  head. 
There  is  now  no  longer  any  moving  mass  of  water  within  the  pipe, 
the  conversion  of  whose  kinetic  energy  into  pressure  energy  serves 
to  maintain  this  high  pressure  and  therefore  the  pipe  begins  to 
contract  and  the  water  to  expand  with  a  consequent  return  to 
normal  static  pressure.  This  process  starts  at  the  reservoir  and 
travels  as  a  wave  to  the  lower  end.  During  this  second  period 
some  of  the  water  stored  within  the  pipe  flows  back  into  the 
reservoir  but  on  account  of  the  inertia  of  this  moving  mass  an 
amount  flows  back  greater  than  the  excess  amount  stored  at  the 
end  of  the  first  period  so  that  the  instant  this  second  wave  reaches 
the  valve  the  pressure  at  that  point  drops  not  only  to  the  normal 
static  pressure  but  below  it.  A  third  period  now  follows  during 
which  a  wave  of  pressure  less  than  static  sweeps  up  the  pipe  to 
the  reservoir.  When  it  reaches  the  reservoir  the  entire  pipe  is 
under  pressure  less  than  static  but  since  all  the  water  is  again  at 
rest  the  pressure  in  /„  immediately  returns  to  the  normal  static 
pressure  due  to  the  head  of  water  in  the  reservoir.  This  starts  a 
fourth  period  marked  by  a  wave  of  normal  static  pressure  moving 
down  the  pipe.  -When  the  valve  is  reached  the  pressure  there  is 
normal  and  for  an  instant  the  conditions  throughout  the  pipe  are 
similar  to  what  they  were  when  the  valve  was  first  closed.  The 
velocity  of  the  water  (and  the  resultant  water  hammer)  is  now, 
however,  somewhat  less  than  it  was  at  the  time  of  closure  because 
of  friction  and  the  imperfect  elasticity  of  the  pipe  and  the  water. 

Instantly  another  cycle  begins  similar  to  the  one  above 
described,  and  then  another,  and  so  on,  each  set  of  waves  succes- 
sively diminishing,  until  finally  the  waves  die  out  from  the  influ- 
ences above  mentioned. 

Equation  (44)  shows  that  for  instantaneous  closure  of  valve 
the  pressure  created  would  be  infinite  if  the  water  were  incom- 
pressible and  the  pipe  were  inelastic.  Instantaneous  closure  is, 
however,  physically  impossible.  To  determine  the  amount  of 
excess  pressure  actually  resulting  from  water  hammer  it  is  neces- 
sary to  take  into  consideration  the  elasticity  of  the  pipe  and  the 
compressibility  of  water.  This  leads  to  a  rather  lengthy  mathe- 


226  HYDRODYNAMICS 

matical  analysis  which  will  here  be  avoided  and  there  will  be  given 
only  the  resulting  workable  equations.  The  following  nomen- 
clature will  ,be  used,  all  units  being  expressed  in  feet  and  seconds 
except  E  and  Ef,  which  are  in  pounds  per  square  inch: 

b  =  thickness  of  pipe  walls; 
D  =  inside  diameter  of  pipe; 

E  =  modulus  of  elasticity  of  pipe  walls  in  tension; 
E'  =  modulus  of  elasticity  of  water  in  compression; 

g  =  acceleration  of  gravity; 

h  =  head  due  to  water  hammer  (in  excess  of  static  head) ; 
H  =  normal  static  head  in  pipe; 
L  =  length  of  pipe  line; 
77  =  time  of  closing  valve; 

v  =  mean  velocity  of  water  in  pipe  before  closure  of  valve; 
vw  =  velocity  of  pressure  wave  along  pipe. 

142.  Formulas  for  Water  Hammer. — In  the  following  dis- 
cussion whenever  the  term  "  pressure  "  is  used  it  is  understood  to 
mean  "  pressure  due  to  water  hammer  "  and  is  the  amount  of 
pressure  in  excess  of  that  due  to  the  normal  static  head. 

If  the  valve  is  closed  instantaneously  the  pressure  in  l\  imme- 
diately rises  to  pmax  and  remains  at  this  value  while  the  pressure 
wave  travels  to  the  reservoir  and  returns.  The  time  required 

nr 

for  the  wave  to  travel  to  the  reservoir  and  back  to  the  valve  is  — -. 

vw 

The  pressure  in  ln  reaches  this  same  pmax  but  remains  at  that 
value  only  for  an  instant.  At  any  intermediate  section,  pm&K 
is  maintained  only  until  the  wave  of  reduced  pressure  reaches  that 

section.     If  the   time  taken  to   completely   close  the  valve   is 

or 
exactly  — ,  pmax  will  occur  only  in  section  l\,  and  will  last  only 

Vw 

for  an  instant,  being  immediately  lowered  by  the  return  of  the 
static  pressure  wave.  If  the  time  of  closing  the  valve  is  greater 

2L 
than  — ,  pmax  will  never  be  attained  since  the  wave  of  reduced 

Vw 

pressure  will  then  have  reached  li  before  the  valve  is  completely 
closed  or  pmax  is  reached. 

Evidently  two  formulas  are  necessary;  one  to  determine  the 
maximum  water  hammer,  when  the  time  of  closure  is  less  than 


FORMULAS   FOR  WATER  HAMMER  227 

Q7" 

—  ,  and  the  other  to  determine  the  ordinary  water  hammer  that 

vw 

or 
occurs  when  the  time  of  closure  is  greater  than  —  ,  as  is  usually 

Vw 

the  case. 

The  same  formula  for  the  determination  of  maximum  water 
hammer  has  been  quite  generally  adopted.  The  general  expres- 
sion is, 

*m»  =  ^,      .........     (45) 

y 
where 

4660 


(46) 


Eb 
Substituting  this  value  in  equation  (45), 


*  f 

(  } 


Eb 

For  steel  pipe  this  reduces  to 

1  A.ZV 

(48) 


Frequently  these  expressions  appear  in  different  forms  but  they 
may  all  be  reduced  to  the  above  forms. 

There  is  no  such  general  agreement  as  to  the  proper  formula  to 
be  used  for  the  determination  of  ordinary  water  hammer  when  T 

2L 

is  greater  than  — .     Many  formulas   have   been   derived,    some 

^w 

giving  results  twice  as  great  as  others.  Certain  assumptions  as 
to  the  manner  of  valve  closure,  the  effect  of  friction  and  the 
manner  in  which  the  waves  are  reflected,  etc.,  must  be  made 
before  any  theoretical  formula  can  be  derived.  It  appears  that 
the  main  reason  for  the  wide  discrepancy  in  results  lies  in  the 
difference  in  these  fundamental  assumptions. 

Assuming  that  the  valve  is  closed  in  such  manner  that  the 
rate  of  rise  in  pressure  will  be  constant  throughout  the  entire 
closure,  OA,  Fig.  133,  represents  the  variation  in  pressure  at  the 


228 


HYDRODYNAMICS 


valve  during  the  time  T,  provided  that  no  return  pressure  wave 

2Z/ 
interferes.     After  a  time,  — ,  however,  the  returning  wave  will 

vw 

reach  the  valve  and  assuming  that  its  intensity  has  been  undi- 
minished  by  friction  or  other  cause,  it  will  exactly  annul  the 
tendency  for  the  pressure  to  increase,  due  to  continued  closing, 
and  as  a  result  the  pressure  remains  constant  during  the  remainder 


FIG.  133. 


of  the  time   T,  as  shown  by  the  horizontal  line  BA'. 
similar  triangles, 


From 


From  which 


g      vw 
2Lv 


(49) 


This  formula  was  first  proposed  by  Professor  Joukovsky  1  of 
Moscow,  Russia,  in  1898  and,  it  is  claimed,  was  substantiated  by  a 
series  of  experiments  which  he  conducted. 

Other  commonly  used  formulas  2  are  as  follows : 


NH 


(Allievi) 


1  N.  JOUKOVSKY:   Trans,  of  Prof.  N.  Joukovsky's  paper  on  Water  Ham- 
mer, by  Boris  Simin;   Journal  American  Waterworks  Association  (1904). 

2  MILTON  M.  WARREN:    Penstock  and  Surge-Tank  Problems.     Trans. 
Amer.  Soc.  Civ.  Eng.,  vol.  79  (1915).     Contains  discussions  of  all  commonly 
used  water  hammer  formulas. 


PROBLEMS  229 

where 


I,     .     .     (Johnson) 

-iV  - 

where 


Lv    and    N  =  2gHT, 

Lv 


(Warren) 


h  =  —m (Mead,  et  al.) 

Joukovsky's  formula  gives  results  twice  as  great  as  Mead's 
formula.  Johnson's  formula  can  readily  be  reduced  to  Allievi's. 
It  may  be  noted  that  these  two  formulas  make  h  vary  with  H, 
which  the  other  formulas  do  not  do.  Justification  for  this  varia- 
tion is  not  apparent. 

Eliminating  Johnson's  formula  (which  reduces  to  Allievi's), 
no  two  of  the  above  formulas  give  results  that  are  at  all  similar 
under  all  conditions.  Discrepancies  of  from  100  to  200  per  cent 
are  possible.  A  comprehensive  and  carefully  conducted  series  of 
experiments  are  necessary  before  any  formula  for  ordinary  water 
hammer  can  be  relied  upon  to  give  trustworthy  results. 

PROBLEMS 

1.  A  jet  1  in.  in  diameter  and  having  a  velocity  of  25  ft.  per  second  strikes 
normally  against  a  fixed,  flat  plate.     Determine  the  pressure  on  the  plate. 

2.  In  Problem  1,  what  would  be  the  pressure  on  the  plate  if  it  were 
moving  with  a  uniform  velocity  of  10  ft.  per  second  in  the  same  direction 
as  the  jet? 

3.  What  should  be  the  velocity  of  the  plate,  in  Problem  2,  if  the  jet  is  to 
perform   the    maximum   amount   of   work?     Determine   the   corresponding 
amount  of  work  in  foot-pounds  per  second. 

4.  A  jet  having  a  diameter  of  2  in.  and  a  velocity  of  40  ft.  per  second 
is  deflected  through  an  angle  of  60°  by  a  fixed,  curved  vane.     Determine  the 
X-  and  F-components  of  the  force  exerted. 

6.  Solve  Problem  4  if  the  vane  is  moving  with  a  velocity  of  25  ft.  per 
second  in  the  same  direction  as  the  jet. 

6.  A  H-in.  nozzle  has  a  coefficient  of  velocity  of  0.97  and  a  coefficient 
of  contraction  of  unity.  The  base  of  the  nozzle  has  a  diameter  of  4  in., 
at  which  point  the  gage  pressure  is  80  Ibs.  per  square  inch.  The  jet  is 


230  HYDRODYNAMICS 

deflected  through  an  angle  of  150°  by  a  double-cusped  vane  that  has  a 
velocity,  in  the  direction  of  the  jet  of  30  ft.  per  second.  What  is  the  pressure 
exerted  on  the  vane  arid  what  is  the  amount  of  work  done  expressed  in 
foot-pounds? 

7.  In  Problem  6  determine  the  velocity  the  vane  must  have  if  the  jet 
is  to  perform  the  maximum  amount  of  work.     What  is  the  maximum  work 
in  foot-pounds? 

8.  If  the  jet,  in  Problem  6,  strikes  a  series  of  vanes  so  arranged  on  the 
periphery  of  a  wheel  that  the  entire  jet  is  deflected  through  an  angle  of  170°, 
what  is  the  maximum  amount  of  work  that  can  be  done? 

9.  A  horizontal  straight  pipe  gradually  reduces  in  diameter  from  12  in.  to  6 
in.    If,  at  the  larger  end,  the  gage  pressure  is  40  Ibs.  per  square  inch  and  the 
velocity  is  10  ft.  per  second,  what  is  the  total  longitudinal  thrust  exerted 
on  the  pipe?     Neglect  friction. 

10.  A  bend  in  a  pipe  line  gradually  reduces  from  24  in.  to  12  in.     The 
deflection  angle  is  60°.     If  at  the  larger  end  the  gage  pressure  is  25  Ibs.  per 
square  inch  and  the  velocity  is  8  ft.  per  second,  determine  the  X-  and  F-com- 
ponents  of  the  dynamic  thrust  exerted  on  the  bend.     Also  determine  the 
X-  and  F-components  of  the  total  thrust  exerted  on  the  bend,  neglecting 
friction. 

11.  A  24-in.  cast-iron  pipe  |  in.  thick  and  6000  ft.  long  discharges  water 
from  a  reservoir  under  a  head  of  80  ft.     What  is  the  pressure  due  to  water 
hammer  resulting  from  the  instantaneous  closure  of  a  valve  at  the  discharge 
end? 

12.  If  the  time  of  closing  the  valve,  in  Problem  11,  is  6  sec.,  determine 
the  resulting  pressure  due  to  water  hammer,  comparing  results  obtained  by 
use  of  formulas  by  Joukovsky,  Johnson,  Warren  and  Mead. 


INDEX 


Absolute  and  gage  pressure,  10 

Absolute  velocity,  214 

Acceleration,  45,  48,  181,  202 

Accuracy  of  computations,  5 

Air  in  pipes,  164,  165 

Algebraic    transformation   of   orifice 

formula,  80 
of  weir  formula,  113 
Allievi  formula  for  water  hammer,  228 
Angle  of  heel,  40,  42 
Angular  velocity,  48 
Approach,  channel  of,   71,   76,   103, 

108 
Approach,  velocity  of,  71,  76,  80,  103, 

108,  112,  120,  128 
Archimedes,  principle  of,  38 
Atmospheric  pressure,  9,  15,  17 
Axis  of  moments,  25 

Backwater,  204,  207 
Barometer,  mercury,  15 

water,  16 

Barr's  experiments,  123 
Base  formula,  52 
Bazin  experiments,  111 

open-channel  formula,  191 

submerged-weir  formula,  121 

weir  formula,  116 
Bends  in  pipes,  loss  of  head.  143,  162 

coefficients  for,  163 

thrust  at,  222 
Bernoulli's  equation,  59,  63,  74,  87, 

141 
Bernoulli's  theorem,  57 

application  to  hydrostatics,  60 

in  practice,  60 
Boiling  point  of  water,  2 
Borda's  mouthpiece,  91 


Bridge  piers,  195 
Broad-crested  weirs,  129 
Buoyancy,  center  of,  39 
Buoyant  force,  39,  40 

Canals,  199 

Capillary  action,  4,  16 

Cast-iron  pipes,  coefficients,  148,  149, 

152,  153 

Center  of  buoyancy,  39,  42 
Center  of  gravity,  31,  39,  42 
Center  of  pressure,  24,  29,  31,  32 

and  center  of  gravity,  31,  32 
Channel  entrance,  131,  202,  203 
Channel  of  approach,  71,  76,  103,  108 
Characteristics  of  jet,  71 
Chezy  formula,  open  channels,  185 

pipes,  146 
Chutes,  129 
Cippoletti  weir,  124 
Circular  channels,  176,  201 

orifice,  71 
Coefficient  of  contraction,  nozzles,  89 

orifices,  77,  78 

tubes,  86,  88,  92,  93 

weirs,  110 
Coefficient  of  discharge,  nozzles,  89 

orifices,  78,  80 

submerged  orifice,  94 

tubes,  86,  93 

Coefficient  of  roughness,  185 
Coefficient  of  velocity,  nozzles,  90 

orifices,  77,  78 

tubes,  86,  93 

weirs,  111 
Coefficients  of  discharge,  nozzles,  90 

orifices,  79 

tubes,  86 


231 


232 


INDEX 


Coefficients,  hose,  150 

open  channels,  187,  190,  191 

pipes,  148,  149,  153.  161,  162,  163 

weirs,  108 

Coefficient,  weir,  111 
Components  of  pressure,  33,  35 
Compound  pipes,  168 
Compressibility  of  water,  3 
Computations,  accuracy  of,  5 
Concrete  pipes,  coefficients,  148,  149, 

153 

Conduits,  135,  199 
Conical  tubes,  88 
Conservation  of  energy,  59 
Converging  tubes,  88 
Continuity  of  discharge,  53 

of  flow,  53,  140,  180,  202 
Contraction,  coefficient  of,  77,  78,  89, 
92,  110 

crest,  weirs,  101 

end,  weirs,  102 

gradual,  pipes,  143,  159 

in  channels,  183,  192 

of  jet,  72,  76,  77,  88,  92 

of  nappe,  103,  108,  110 

sudden,  pipes,  143,  158 

suppressed,  84,  102 

surface,  weirs,  102 

vertical,  weirs,  103 
Converging  tubes,  88 
Crest  contraction,  101 
Critical  velocity,  136 
Cross-section,  most  efficient,  198 
Curves,  backwater,  205 

in  open  channels,  183,  196 

in  pipes,  143,  162 
Curved  surface,  pressure  on,  33 

Dams,  coefficients  for,  127 

pressure  against,  33,  34 
Darcy,  experiments  on  Pitot  tube,  65, 
66 

modification  of  Pitot  tube,  68 

pipe  formula,  147 
Density  of  water,  2 
Depth  of  flotation,  39 
Differential  gage,  18 
Direction  of  resultant  pressure,  7 


Discharge,  52 

continuity  of,  53,  140,  180 

head  lost  at,  143,  155,  157,  183 

under  falling  head,  96 
Discharge  coefficient,  nozzles,  89 

orifices,  78 

tubes,  86,  93 

Discussion  of  open-channel  formulas, 
191 

of  pipe  formulas,  154 

of  weir  formulas,  118 
Distance,  unit  of,  1 
Distilled  water,  properties  of,  2 
Distribution  of  velocities,   108,  138, 

176 

Diverging  tube,  90 
Divided  flow,  169,  206 
Dynamic  force,  214-229 

Efficient  channel  section,  198 
Elasticity  of  water,  3 
Elevation  and  air  pressure,  9 
Emptying  vessel,  97 
End  contractions,  weirs,  102 

suppressed,  102 
Energy  and  head,  54 

kinetic,  54,  57 

of  position,  55,  57 

of  water  in  channel,  109,  179 

of  water  in  pipe,  139 

per  pound  of  water,  57,  140 

potential,  55,  57 

pressure,  56 

English  system  of  units,  1 
Enlargement  of  section,  channels,  193 

pipes,  143 
Entrance  losses,  channels,  183 

pipes,  143,  155,  156 
Entrance  to  channels,  131 
Equation  of  continuity,  53,  181,  202 

Falling  head,  discharge  under,  96 
Falls,  117 
Floating  bodies,  38 

stability  of,  39 
Flow,  continuity  of,  53,  140,  180,  202 

in  open  channels,  176-209 

in  rivers,  196 


INDEX 


233 


Flow,  non-uniform,  53,  201 

over  weirs,  101-134 

steady,  53 

stream  line,  54,  136 

through  nozzles,  89 

through  orifices,  71-99 

through  pipes,  135-175 

through  tubes,  85-93 

turbulent,  54,  136 

uniform,  53 
Fluids,  definition,  1 
Flumes,  199 
Force,  acting  on  mass,  212,  213 

exerted  on  pipe,  220-223 

exerted  on  vane,  213-218 
Francis  coefficient  for  weirs,  111 

experiments,  111 

formula  for  end  contractions,  112 

weir  formula,  115 
Free  surface  of  liquid,  9 
Freezing  temperature  of  water,  2 
Friction,  52 

and  distribution  of  velocities,  108, 
138,  176 

Bernoulli's  equation  with,  59,  141 

coefficients  for  hose,  150 

coefficients  for  open  channels,  187, 
190,  191 

coefficients  for  pipes,  148,  149,  153 

formulas    for  open  channels,  185- 
191 

formulas  for  pipes,  146-154 

independent  of  pressure,  145 

in  open  channels,  176,  184 

in  orifices,  76,  77 

in  pipes,  138-146 

in  Venturi  meter,  64 

of  flowing  water,  59,  138,  176 

over  weirs,  108 
Fteley  and  Stearns,  experiments,  111 

weir  formula,  116 
Fundamental  orifice  formula,  73 

Gage,  differential,  18 
hook,  131 
mercury,  17 
oil,  19 
water  stage,  132 


Gage  pressure,  10,  16 
Ganguillet  and  Kutter,  185 
Gases,  1,  9 
Gates,  95,  195 

and  valves,  143,  162 

coefficients  for,  96 

pressure  on,  26,  28,  32 
Grade  line,  hydraulic,  144,  154,  164, 

184,  191 

Gradient,  hydraulic,   144,   154,   164, 
184, 191 

Hazen- Williams  formula,  150 
Head,  definition  of,  54 

energy  and,  54 

lost  at  bends  or  curves,  143,  162, 

183,  196 

lost  at  discharge,  143,  155,  157,  194 

lost  at  entrance,  143,  155,  156,  193 

lost  at  obstructions,  143.  162,  195 

lost  by  contractions,  143,  158,  192 

lost  by  enlargements,  143,  160,  193 

lost  by  friction,  59 

lost  by  friction  in  open  channels, 
181,  183,  184 

lost  by  friction  in  pipes,  140,  'l43, 
144 

lost  in  nozzles,  90 

lost  in  orifices,  81 

lost  in  tubes,  87 

measurement  of,  131 

pressure,  13,  16,  56 

velocity,  55 
Height  of  weir,  103 
Herschel,  Venturi  meter,  61 
Hook  gage,  4,  131 
Horizontal  and  vertical  components 

of  pressure,  32,  34 
Horizontal  orifice,  71,  75 
Horton's  values  of  "n,"  187 

weir  coefficients,  127 
Hose,  150 

coefficients,  150 
Hydraulic  grade  line,  144,  154,  164, 

184,  191 

gradient,  144,  154,  164,  184,  191 
jack,  13 
radius.  135,  176 


234 


INDEX 


Hydraulics,  definition,  J 

or  rivers,  196 
Hydrodynamics,  211-230 

definition,  1 

fundamental  principles,  211 
Hydrokinetics,  1,  51-210 
Hydrostatic  pressure,  7-37 
Hydrostatics,  1,  7-44 

Immersed  and  floating  bodies,  38-44 
Impurities  in  water,  2 
Incomplete  contractions,  84,  103 
Intensity  of  pressure,  7,  8,  10,  30,  35 
Inversion  of  jet,  73 
Inward-projecting  tubes,  93 
Irregular  sections  of  channels,  197 

Jack,  hydraulic,  13 

Jet,  characteristics  of,  71 

contraction  of,  72,  76,  77,  88,  92 

definition,  71 

deflected  by  curved  vane,  216 

force  of,  213-218 

forms  assumed  by,  72 

impinging  on  fixed  plate,  214 

inTpinging  on  moving  plate,  215 

inversion  of,  73 

path  of,  82 

pressure  in,  72 

velocity  with  respect  to  vane,  217 

work  done  by,  218 
Johnson  water-hammer  formula,  229 
Joukovsky,     water-hammer    experi- 
ments, 228 

formula,  228 

Kinetic  energy,  54,  57 

in  channel  of  approach,  109 
in  open  channels,  109,  179 
in  pipes,  139 

King  pipe  formula,  152 

submerged-weir  formula,  121 
weir  formula,  117 

Kutter  formula,  185 
table  for  solving,  186 

Lea,  investigation  of  pipes,  152 
Linear  acceleration,  45 


Liquid,  buoyant  force  of,  38 

definition,  1 

free  surface  of,  9 

in  motion,  51 

intensity  of  pressure  within,  7,  8 

relative  equilibrium,  45 

rotating,  47 

surface  tension,  4 

uniformly  accelerated,  45 

water  most  common,  51 
Long  pipes,  166 

Loss  of  head,  at  discharge,  143,  155, 
157,  183,  194 

at  entrance,  143,  155,  156,  183,  194 

due  to  bends  or  curves,  143,  162, 
183,  196 

due  to  contractions,  143,  158,  192 

due  to  enlargements,  143,  160,  193 

due  to  friction  in  channels,  181,  184 

due  to  friction  in  pipes,  140-146 

due  to  obstructions,  143,  162,  195 

in  nozzles,  90 

in  open  channels,  181 

in  orifices,  81 

hi  pipes,  140 

in  tubes,  87 

Manning  formula,  188 
Mead  water-hammer  formula,  229 
Mean  velocity  over  weirs,  107 
Measurement  of  head,  131 
Mercury  barometer,  15 

gage,  17 
Metacenter,  40 
Metacentric  height,  40 
Meter,  Venturi,  61 
Modifications   of   fundamental   weir 

formula,  112 
Modulus  of  elasticity,  3 
Moment  of  inertia,  25 
Momentum,  definition,  211 
Motion,  stream  line,  54,  136 

turbulent,  54,  136 
Mouthpiece,  Borda's,  91 

Nappe,  101 

contracted  section  of,  103 
contraction  of,  108,  110 
depth  of  mean  velocity  in,  108 


INDEX 


235 


Nappe,  pressure  in,  104 

velocities  in,  104,  107,  111 
Negative  pressure,  10 
Newton's  laws  of  motion,  211 
Non-uniform  flow,  53,  201 
Nozzles,  89 
Numerical  computations,  5 

Obstructions  in  channels,  183,  195 

in  pipes,  143,  162 
Oil-differential  gage,  19 
Open-channel  friction  formulas,  185- 
191 

Bazin  formula,  191 

Chezy  formula,  185 

Kutter  formula,  185,  188 

Manning  formula,  188 
Open  channels,  backwater  in,  204 

bends  or  curves  in,  183,  196 

contractions  in  section,  183,  192 

description  and  definition,  176 

discussion  of  formulas,  191 

divided  flow  in,  206 

efficient  section  for,  198 

energy  of  water  in,  109,  179 

enlargement  in  section,  183,  193 

entrance  to,  131,  183 

friction  in,  176 

hydraulic  gradient  of,  184,  191 

hydraulic  radius  of,  176 

loss  of  head  in,  181 

non-uniform  flow,  201 

obstructions  in,  183,  195 

water  surface  of,  184,  191 

wetted  perimeter  of,  176 
Orifice,  circular,  71 

coefficients,  76 

definition  and  description,  71 

discharge  under  falling  head,  96 

flow  through,  71-85,  93-98 

fundamental  formula,  73 

head  lost  in,  81 

horizontal,  71,  75 

partially  submerged,  94 

rectangular,  71,  82 

square,  71 

submerged,  93 

suppressed  contraction,  84 


Orifice,  under  low  heads,  82 
vertical,  71 
with  velocity  of  approach,  80 

Parabolic  weir,  101 

Partially  submerged  orifices,  94 

Pascal's  law,  8 

Path  of  jet,  82 

Perfect  liquid,  4 

Piezometer  tubes,  5,  16 

Pipe-friction  formulas,  146-154 

Chezy  formula,  146 

Dracy  formula,  147 

Hazen- Williams  formula,  150 

King  formula,  152 

non-turbulent  flow,  154 
Pipes,  bends  or  curves  in,  143,  162 

branching,  166 

coefficients,  148,  149,  151,  153 

compound,  168 

continuity  of  flow,  140 

contractions  in,  143,  158 

critical  velocity  in,  136 

description  and  definition,  135 

discharge  from,  143,  155,  157 

divided  flow  in,  169 

energy  of  water  in,  139 

enlargements  in,  143,  160 

entrance  to,  143,  155,  156 

flow  through,  135-175 

force  exerted  upon,  220 

friction  coefficients,  148,  149,  153 

friction  in,  138,  144 

hydraulic  gradient,  144,  154,  164 

hydraulic  radius  of,  135 

loss  of  head  in,  140 

more  than  one  diameter,  142,  172 

obstructions  in,  143,  162 

special  problems,  165 

tensile  stress  in  walls,  35 

thrust  at  bends,  222 

viscosity,  effect  on  friction,  145 

water  hammer  in,  223 

wetted  perimeter  of,  135 
Piston,  13,  56 
Pitot  tube,  65-69 

Plate,  jet  impinging  against,  214,  215 
Potential  energy,  55,  57 


236 


INDEX 


Pressure,  absolute  and  gage,  10 

atmospheric,  9,  15 

center  of,  24,  29,  31 

energy,  56,  62 

head,  13,  16,  56 

intensity  of,  7,  8,  10,  30,  32,  35 

negative,  10 

on  curved  surfaces,  33,  35 

on  plane  areas,  23 

on  surfaces,  23-37 

relative,  10 

transmission  of,  13 

vapor,  14,  164 
Principle  of  Archimedes,  38 
Principles  of  equilibrium,  8 
Principles  of  hydrokinetics,  51-70 
Principles  of  hydrostatic  pressure,  7- 

22 
Pumps,  suction,  20 

Radius,  hydraulic,  135,  176 

of  bends  in  pipes,  163 
Reaction  of  jet,  213 
Rectangular  channels,  176,  199 

orifice,  71,  82 

weir,  101 

Relative  and  absolute  velocities,  214 
Relative  equilibrium  of  liquids,  45-50 
Relative  pressure,  10 
Resultant  pressure,  direction  of,  7,  33 

position  of,  32 
Rivers,  hydraulics  of,  196 
Rotating  vessel,  47 

Salt  water,  3 

Scow,  stability  of,  42 

Section  of  greatest  efficiency,  198 

Semicircular  channel,  198 

Sewers,  135,  176 

Ships,  stability  of,  40 

Siphon,  20 

Skin  of  water  surface,  4 

Slope,  "s,"  145,  184 

Special  pipe  problems,  165 

Specific  gravity  of  mercury,  17 

of  water,  2 
Square  orifice,  71 
Stability  of  scow,  42 


Stability  of  ship,  40 
Standard  short  tube,  85 
Steady  flow,  53 
Stream  line  motion,  54,  136 
Submerged  bodies,  38 

orifices,  93 

weirs,  95,  118,  120,  195 
Suction  pumps,  20 
Sudbury  conduit,  177 
Summit  of  pipe,  164,  165 

of  siphon,  20 

Suppressed  contraction,  orifices,  84 
Suppressed  weirs,  102 
Suppression  of  contraction,  84,  102 
Surface  contraction,  102 
Surfaces,  pressure  on,  23-37 
Surface  tension,  4,  177 

Tangential  stress  of  water,  1,  4,  8 
Temperature,  2 

effect  on  critical  velocity,  138 

effect  on  viscosity,  138 

effect  on  weight  of  water,  2 
Tensile  stress  in  pipe  walls,  35 
Theoretical  discharge  over  weirs,  105 

velocity  over  weirs,  107 

velocity  through  orifices,  74 
Thickness  of  pipes  required,  35 
Thompson's  experiments,  123 
Time  of  emptying  vessel,  97 
Time,  unit  of,  1 
Torricelli's  theorem,  76 
Transmission  of  pressure,  13 
Trapezoidal  channels,  176,  199 

weirs,  123 

Triangular  weirs,  121 
Tube,  definition  of,  71 
Tubes,  capillary  action  in,  4 

converging,  88 

diverging,  90 

immersed  in  flowing  water,  65 

standard  short,  85 
Turbulent  motion,  54,  136 

Uniform  flow,  53 
Units  in  hydraulics,  1 
Unsteady  flow,  53 
U-tube,  17 


INDEX 


237 


Vacuum,  10,  17 

amount  of,  10 

perfect,  10 
Vane,  jet  deflected  by,  215,  216 

fixed  curved,  215 

moving  curved,  216 

work  done  on,  218 
Vapor  pressure,  14,  164 

variation  with  temperature,  15 
Velocities,  distribution  of,  138,  176 

in  vertical,  177 
Velocity,  absolute  and  relative,  214 

at  any  depth,  over  weirs,  104 

critical,  136 

curves  of,  177 

from  orifices,  74 

head,  55,  81 

of  approach,  orifice,  71,  75,  76,  78, 
80 

weir,  103,  108,  112,  114,  120,  128 
Velocity  coefficient,  nozzles,  90 

orifices,  77,  78 

tubes,  86,  93 

weirs,  111 
Vena  contracta,  72 
Venturi  meter,  61 
Vertical  contraction,  103 
Vertical  jets,  75 
Vertical  orifice,  71 
Vertical  velocity  curves,  177 
Vessel,  rotating,  47 

time  of  emptying,  96 

with  constant  acceleration,  45 
Viscosity,  4,  137,  143 

Warren  water-hammer  formula,  229 
Water  barometer,  16 

boiling  point  of,  2 

compressibility  of,  3 

dynamic  force,  214-229 

freezing  point  of,  2 

hammer,  223 

maximum  density  of,  2 

physical  properties  of,  2-5 


Water,  weight  of,  2 

Weight  of  submerged  bodies,  38 

of  mercury,  17 

of  water,  2 

unit  of,  1 
Weir,  broad  crested,  129 

channel  of  approach,  71,  76,  103, 
108 

Cippoletti,  124 

coefficient,  111,  127 

coefficients,  108,  110,  111 

conditions  for  accurate  use,  132 

definition  and  description,  101 

discharge  formulas  for  sharp  crest, 
115 

discussion  of  formulas,  118 

experiments,  111,  115 

formula  for  mean  velocity,  105 

fundamental  formulas,  83,  107,  112 

head  measurement,  131 

height  of,  103 

modification   of  fundamental   for- 
mula, 112 

not  sharp-crested,  101,  103,  125 

range  of  accuracy,  132 

rectangular,  101 

sections,  126 

sharp-crested,  101 

submerged,  95,  118 

theoretical  formula  for  discharge, 
105 

transformation  of  formula,  113 

trapezoidal,  101,  123 

triangular,  101,  121 

velocity  of  approach  to,  103 

with  end  contractions,  102,  112 

with  end  contractions  suppressed, 

102 

Wetted  perimeter,  135,  176 
Wooden  pipes,  coefficients,  148,  149, 

153 
Work  done  on  vane,  218 

on  mass  of  water,  58 


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